OP Malhotra Class 12 Maths Solutions Chapter 14 Indefinite Integral 2 Exercise 14 (E)

Question 1. Evaluate: \( \int e^x(\cot x+\log \sin x) d x \)
Answer:
Let \( I = \int e^x (\cot x + \log \sin x) dx \)
We know the formula: \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \).
Here, let \( f(x) = \log \sin x \).
Then, \( f'(x) = \frac{d}{dx}(\log \sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x \).
So, the given integral is in the form \( \int e^x (f'(x) + f(x)) dx \).
Therefore, \( I = e^x \log \sin x + C \). This rule simplifies integrals very quickly.
In simple words: This problem uses a special integration rule. If you have `e` to the power of `x` multiplied by a function plus its derivative, the answer is just `e` to the power of `x` times the original function.

🎯 Exam Tip: Always look for the \( \int e^x (f(x) + f'(x)) dx \) pattern when \( e^x \) is present in the integrand, as it's a common shortcut that saves time.

Question 2. \( \int e^{2 x}(-\sin x+2 \cos x) d x \)
Answer:
Let \( I = \int e^{2x} (-\sin x + 2 \cos x) dx \)
We use the integration by parts method. Let's consider integrating \( \int e^{2x} (2 \cos x) dx \) first.
Let \( u = 2 \cos x \) and \( dv = e^{2x} dx \).
Then \( du = -2 \sin x dx \) and \( v = \frac{e^{2x}}{2} \).
Using the formula \( \int u dv = uv - \int v du \):
\( \int e^{2x} (2 \cos x) dx = 2 \cos x \cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} (-2 \sin x) dx \)
\( = e^{2x} \cos x + \int e^{2x} \sin x dx \).
Now, substitute this back into the original integral for \( I \):
\( I = \int e^{2x} (-\sin x) dx + \left( e^{2x} \cos x + \int e^{2x} \sin x dx \right) \)
\( I = -\int e^{2x} \sin x dx + e^{2x} \cos x + \int e^{2x} \sin x dx \)
The integral terms cancel out.
\( I = e^{2x} \cos x + C \). This pattern highlights how careful selection for integration by parts can simplify complex expressions.
In simple words: We used a method called integration by parts. When we split the problem and did one part, it magically canceled out the other part, leaving a simple answer.

🎯 Exam Tip: When integrating forms like \( \int e^{ax}(A \cos bx + B \sin bx) dx \), recognize that integration by parts often leads to the original integral reappearing, which can then be solved algebraically, or look for the specific form \( \int e^{ax} [a f(x) + f'(x)] dx \).

Question 3. \( \int e^x(\tan x+\log \sec x) d x \text { or } \int e^x(\tan x-\log \cos x) d x \)
Answer:
Let \( I = \int e^x (\tan x + \log \sec x) dx \)
We use the formula \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \).
Here, let \( f(x) = \log \sec x \).
Then, \( f'(x) = \frac{d}{dx}(\log \sec x) = \frac{1}{\sec x} \cdot (\sec x \tan x) = \tan x \).
So, the integral is in the form \( \int e^x (f(x) + f'(x)) dx \).
Therefore, \( I = e^x \log \sec x + C \). This integral form is a direct application of a fundamental integration property.
In simple words: This problem fits a special rule where if you have `e` to the power of `x` multiplied by a function and its derivative, the answer is `e` to the power of `x` times that function.

🎯 Exam Tip: Remember that \( \log \sec x = -\log \cos x \). This means both forms of the question are equivalent, and the solution path remains the same using the \( \int e^x (f(x) + f'(x)) dx \) formula.

Question 4. \( \int\{\sin (\log x)+\cos (\log x)\} d x \)
Answer:
Let \( I = \int [\sin(\log x) + \cos(\log x)] dx \)
Let's use a substitution. Let \( \log x = t \).
Then \( x = e^t \). So, \( dx = e^t dt \).
Substitute these into the integral:
\( I = \int (\sin t + \cos t) e^t dt \)
This is in the form \( \int e^t (f(t) + f'(t)) dt = e^t f(t) + C \).
Here, let \( f(t) = \sin t \). Then \( f'(t) = \cos t \).
So, \( I = e^t \sin t + C \). This powerful substitution simplifies the problem significantly.
Now, substitute back \( t = \log x \):
\( I = e^{\log x} \sin(\log x) + C \)
Since \( e^{\log x} = x \),
\( I = x \sin(\log x) + C \).
In simple words: We changed `log x` to `t` to make the problem simpler. After solving, we changed `t` back to `log x` to get the final answer.

🎯 Exam Tip: When integrating expressions involving \( \log x \) or \( \sin(\log x) \), a substitution like \( \log x = t \) (which implies \( x = e^t \) and \( dx = e^t dt \)) often transforms the integral into a standard form involving \( e^t(f(t) + f'(t)) \).

Question 5. \( \int\left\{\tan (\log x)+\sec ^2(\log x)\right\} d x \)
Answer:
Let \( I = \int [\tan(\log x) + \sec^2(\log x)] dx \)
Let's use a substitution. Let \( \log x = t \).
Then \( x = e^t \). So, \( dx = e^t dt \).
Substitute these into the integral:
\( I = \int (\tan t + \sec^2 t) e^t dt \)
This is in the form \( \int e^t (f(t) + f'(t)) dt = e^t f(t) + C \).
Here, let \( f(t) = \tan t \). Then \( f'(t) = \sec^2 t \).
So, \( I = e^t \tan t + C \). Recognizing this pattern is key for these types of integrals.
Now, substitute back \( t = \log x \):
\( I = e^{\log x} \tan(\log x) + C \)
Since \( e^{\log x} = x \),
\( I = x \tan(\log x) + C \).
In simple words: We used a simple trick: replace `log x` with `t`. Then the problem became a known type that's easy to solve, and we put `log x` back at the end.

🎯 Exam Tip: Whenever you see functions of \( \log x \) in an integrand, especially when \( e^x \) is implicitly or explicitly involved, consider the substitution \( t = \log x \) to simplify the expression into the standard \( \int e^t(f(t) + f'(t)) dt \) form.

Question 6. \( \int e^x\left(\sin ^{-1} x+\frac{1}{\sqrt{1-x^2}}\right) d x \)
Answer:
Let \( I = \int e^x \left( \sin^{-1} x + \frac{1}{\sqrt{1-x^2}} \right) dx \)
We use the formula \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \).
Here, let \( f(x) = \sin^{-1} x \).
Then, \( f'(x) = \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \).
So, the integral is directly in the form \( \int e^x (f(x) + f'(x)) dx \).
Therefore, \( I = e^x \sin^{-1} x + C \). This is a straightforward application of a common integration identity.
In simple words: This integral is a perfect match for a known rule. We identify a function and its derivative, and the solution simply becomes `e` to the power of `x` times that function.

🎯 Exam Tip: Memorize the derivatives of inverse trigonometric functions, especially \( \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \), as they frequently appear in combination with \( e^x \) for this specific integration pattern.

Question 7. \( \int e^x[\sec x+\log (\sec x+\tan x)] d x \)
Answer:
Let \( I = \int e^x [\sec x + \log (\sec x + \tan x)] dx \)
We use the formula \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \).
Here, let \( f(x) = \log (\sec x + \tan x) \).
To find \( f'(x) \), we apply the chain rule:
\( f'(x) = \frac{d}{dx} (\log (\sec x + \tan x)) = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx} (\sec x + \tan x) \)
\( f'(x) = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x) \)
\( f'(x) = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} \)
\( f'(x) = \sec x \).
So, the integral is directly in the form \( \int e^x (f(x) + f'(x)) dx \).
Therefore, \( I = e^x \log (\sec x + \tan x) + C \). This integral helps in understanding combined functions in calculus.
In simple words: We found that one part of the function was the derivative of the other part. Because `e` to the power `x` was also there, we used a special rule to get the answer.

🎯 Exam Tip: Recognize that \( \frac{d}{dx} (\log |\sec x + \tan x|) = \sec x \). This is a standard derivative identity that is crucial for solving integrals of this type quickly.

Question 8. \( \int e^x \frac{1-\sin x}{1-\cos x} d x \)
Answer:
Let \( I = \int e^x \frac{1-\sin x}{1-\cos x} d x \)
We need to simplify the fraction using half-angle formulas.
Recall: \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \) and \( 1 - \cos x = 2 \sin^2 \frac{x}{2} \).
Substitute these into the fraction:
\( \frac{1-\sin x}{1-\cos x} = \frac{1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \)
Separate the terms:
\( = \frac{1}{2 \sin^2 \frac{x}{2}} - \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \)
\( = \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2} - \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \)
\( = \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2} - \cot \frac{x}{2} \). This transformation using trigonometric identities is a common strategy in integration.
Now the integral becomes:
\( I = \int e^x \left( -\cot \frac{x}{2} + \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2} \right) dx \)
This is in the form \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \).
Here, let \( f(x) = -\cot \frac{x}{2} \).
Then \( f'(x) = -\frac{d}{dx} \left( \cot \frac{x}{2} \right) = - \left( -\operatorname{cosec}^2 \frac{x}{2} \right) \cdot \frac{1}{2} = \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2} \).
So, \( I = e^x \left( -\cot \frac{x}{2} \right) + C = -e^x \cot \frac{x}{2} + C \).
In simple words: We used half-angle formulas to change the fraction into a simpler form. After that, the problem matched a special integration rule that made finding the answer easy.

🎯 Exam Tip: Always try to simplify complex trigonometric fractions in an integral using identities (especially half-angle or double-angle formulas) to see if they can be reduced to the \( f(x) + f'(x) \) form for \( e^x \).

Question 9. \( \int e^x \frac{2-\sin 2 x}{1-\cos 2 x} d x \)
Answer:
Let \( I = \int e^x \frac{2-\sin 2x}{1-\cos 2x} d x \)
We need to simplify the fraction using double-angle formulas.
Recall: \( \sin 2x = 2 \sin x \cos x \) and \( 1 - \cos 2x = 2 \sin^2 x \).
Substitute these into the fraction:
\( \frac{2-\sin 2x}{1-\cos 2x} = \frac{2 - 2 \sin x \cos x}{2 \sin^2 x} \)
Separate the terms:
\( = \frac{2}{2 \sin^2 x} - \frac{2 \sin x \cos x}{2 \sin^2 x} \)
\( = \frac{1}{\sin^2 x} - \frac{\cos x}{\sin x} \)
\( = \operatorname{cosec}^2 x - \cot x \). This simplification is crucial for applying the standard integration form.
Now the integral becomes:
\( I = \int e^x (\operatorname{cosec}^2 x - \cot x) dx \)
Rearrange the terms to match the formula \( \int e^x (f(x) + f'(x)) dx \):
\( I = \int e^x (-\cot x + \operatorname{cosec}^2 x) dx \)
Here, let \( f(x) = -\cot x \).
Then \( f'(x) = -\frac{d}{dx}(\cot x) = -(-\operatorname{cosec}^2 x) = \operatorname{cosec}^2 x \).
So, the integral is in the form \( \int e^x (f(x) + f'(x)) dx \).
Therefore, \( I = e^x (-\cot x) + C = -e^x \cot x + C \).
In simple words: We used angle formulas to make the fraction simpler. Then, we found a function and its derivative inside the integral, which allowed us to use a special rule to quickly find the answer.

🎯 Exam Tip: Always look for opportunities to simplify the integrand using trigonometric identities (like double-angle formulas) to transform it into the \( e^x(f(x) + f'(x)) \) form, which is a common strategy in these types of problems.

Question 10. \( \int\left(\log (\log x)+\frac{1}{\log x}\right) d x \)
Answer:
Let \( I = \int \left( \log (\log x) + \frac{1}{\log x} \right) dx \)
Let's use a substitution. Let \( \log x = t \).
Then \( x = e^t \). So, \( dx = e^t dt \).
Substitute these into the integral:
\( I = \int \left( \log t + \frac{1}{t} \right) e^t dt \)
This is in the form \( \int e^t (f(t) + f'(t)) dt = e^t f(t) + C \).
Here, let \( f(t) = \log t \). Then \( f'(t) = \frac{1}{t} \).
So, \( I = e^t \log t + C \). This substitution effectively converts the complex expression into a standard integrable form.
Now, substitute back \( t = \log x \):
\( I = e^{\log x} \log(\log x) + C \)
Since \( e^{\log x} = x \),
\( I = x \log(\log x) + C \).
In simple words: We made the problem easier by changing `log x` to `t`. This made the integral fit a known pattern, and after solving, we put `log x` back to get the final answer.

🎯 Exam Tip: For integrals containing \( \log (\log x) \) or \( \frac{1}{\log x} \), the substitution \( t = \log x \) is almost always the correct first step, leading to the recognizable \( \int e^t(f(t) + f'(t)) dt \) form.

Question 11. \( \int \frac{\log x}{(1+\log x)^2} d x \)
Answer:
Let \( I = \int \frac{\log x}{(1+\log x)^2} d x \)
Let's use a substitution. Let \( \log x = t \).
Then \( x = e^t \). So, \( dx = e^t dt \).
Substitute these into the integral:
\( I = \int \frac{t}{(1+t)^2} e^t dt \)
Now, rewrite the fraction \( \frac{t}{(1+t)^2} \) to fit the \( f(t) + f'(t) \) form:
\( \frac{t}{(1+t)^2} = \frac{(1+t) - 1}{(1+t)^2} = \frac{1}{1+t} - \frac{1}{(1+t)^2} \). This algebraic manipulation is often necessary before applying the exponential rule.
So, the integral becomes:
\( I = \int e^t \left( \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt \)
This is in the form \( \int e^t (f(t) + f'(t)) dt = e^t f(t) + C \).
Here, let \( f(t) = \frac{1}{1+t} \).
Then \( f'(t) = \frac{d}{dt} \left( (1+t)^{-1} \right) = -1 (1+t)^{-2} \cdot 1 = -\frac{1}{(1+t)^2} \).
So, \( I = e^t \cdot \frac{1}{1+t} + C \).
Now, substitute back \( t = \log x \):
\( I = e^{\log x} \cdot \frac{1}{1+\log x} + C \)
Since \( e^{\log x} = x \),
\( I = \frac{x}{1+\log x} + C \).
In simple words: We used a substitution to simplify the integral. Then, we rewrote the fraction so it matched a special pattern, which allowed us to easily solve the integral and substitute back the original variable.

🎯 Exam Tip: For integrals of the form \( \int e^x \frac{f(x)}{(1+f(x))^2} dx \) or similar, try to manipulate the numerator to create terms that fit \( \int e^t(f(t) + f'(t)) dt \) after substitution.

Question 12. \( \int \frac{1-x}{x^2} e^x d x \)
Answer:
Let \( I = \int \frac{1-x}{x^2} e^x d x \)
First, separate the terms in the fraction:
\( \frac{1-x}{x^2} = \frac{1}{x^2} - \frac{x}{x^2} = \frac{1}{x^2} - \frac{1}{x} \). This algebraic simplification makes the function recognizable.
So, the integral becomes:
\( I = \int e^x \left( \frac{1}{x^2} - \frac{1}{x} \right) dx \)
Rearrange the terms to fit the formula \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \):
\( I = \int e^x \left( -\frac{1}{x} + \frac{1}{x^2} \right) dx \)
Here, let \( f(x) = -\frac{1}{x} \).
Then \( f'(x) = \frac{d}{dx} \left( -x^{-1} \right) = -(-1)x^{-2} = \frac{1}{x^2} \).
So, the integral is in the form \( \int e^x (f(x) + f'(x)) dx \).
Therefore, \( I = e^x \left( -\frac{1}{x} \right) + C = -\frac{e^x}{x} + C \).
In simple words: We broke the fraction into two parts. Then, we saw that one part was the function and the other was its derivative, which let us use a special rule to find the answer quickly.

🎯 Exam Tip: When faced with a rational function multiplied by \( e^x \), try to split the rational function into two terms, one of which is the derivative of the other, to apply the \( \int e^x(f(x) + f'(x)) dx \) formula.

Question 13. \( \int e^{2 x}\left(\frac{2 x-1}{4 x^2}\right) d x \)
Answer:
Let \( I = \int e^{2x} \left( \frac{2x-1}{4x^2} \right) d x \)
First, separate the terms in the fraction:
\( \frac{2x-1}{4x^2} = \frac{2x}{4x^2} - \frac{1}{4x^2} = \frac{1}{2x} - \frac{1}{4x^2} \). This step helps in identifying suitable functions.
So, the integral becomes:
\( I = \int e^{2x} \left( \frac{1}{2x} - \frac{1}{4x^2} \right) dx \)
This integral is of the form \( \int e^{ax} [a f(x) + f'(x)] dx = e^{ax} f(x) + C \).
Here, \( a=2 \). Let \( f(x) = \frac{1}{4x} \).
Then \( a f(x) = 2 \cdot \frac{1}{4x} = \frac{1}{2x} \).
And \( f'(x) = \frac{d}{dx} \left( \frac{1}{4x} \right) = \frac{d}{dx} \left( \frac{1}{4} x^{-1} \right) = \frac{1}{4} (-1) x^{-2} = -\frac{1}{4x^2} \).
So, the integral is \( \int e^{2x} [a f(x) + f'(x)] dx \).
Therefore, \( I = e^{2x} \left( \frac{1}{4x} \right) + C = \frac{e^{2x}}{4x} + C \). This rule makes solving such exponential-rational integrals efficient.
In simple words: We broke the fraction into two parts and identified a specific pattern. For this pattern, when `e` to the power of `ax` is multiplied by (`a` times a function + the function's derivative), the answer is just `e` to the power of `ax` times that function.

🎯 Exam Tip: For integrals of the form \( \int e^{ax} g(x) dx \), try to rewrite \( g(x) \) as \( a f(x) + f'(x) \). This transformation is key for applying the special rule \( e^{ax} f(x) + C \).

Question 14. \( \int e^x\left(\frac{1}{x^2}-\frac{2}{x^3}\right) d x \)
Answer:
Let \( I = \int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) d x \)
This integral is directly in the form \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \).
Here, let \( f(x) = \frac{1}{x^2} \).
Then \( f'(x) = \frac{d}{dx} \left( x^{-2} \right) = -2x^{-3} = -\frac{2}{x^3} \).
So, the integral fits the pattern where \( f(x) = \frac{1}{x^2} \) and \( f'(x) = -\frac{2}{x^3} \).
Therefore, \( I = e^x \cdot \frac{1}{x^2} + C = \frac{e^x}{x^2} + C \). Recognizing this pattern simplifies the integration process considerably.
In simple words: We found that one part of the function was the derivative of the other part. Since `e` to the power `x` was also present, we used a special rule to find the answer directly.

🎯 Exam Tip: Keep an eye out for terms like \( \frac{1}{x^n} \) and \( \frac{-n}{x^{n+1}} \) when \( e^x \) is involved, as they often represent \( f(x) \) and \( f'(x) \) respectively, simplifying the integral directly.

Question 15. \( \int \frac{x+\sin x}{1+\cos x} d x \)
Answer:
Let \( I = \int \frac{x+\sin x}{1+\cos x} d x \)
First, simplify the fraction using half-angle formulas.
Recall: \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \) and \( 1 + \cos x = 2 \cos^2 \frac{x}{2} \).
Substitute these into the fraction:
\( \frac{x+\sin x}{1+\cos x} = \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \)
Separate the terms:
\( = \frac{x}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \)
\( = \frac{1}{2} x \sec^2 \frac{x}{2} + \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} \)
\( = \frac{1}{2} x \sec^2 \frac{x}{2} + \tan \frac{x}{2} \). This simplification is essential for a direct integration.
Now the integral becomes:
\( I = \int \left( \frac{1}{2} x \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) dx \)
Let's consider the term \( \int \frac{1}{2} x \sec^2 \frac{x}{2} dx \) and apply integration by parts to it.
Let \( u = x \) and \( dv = \frac{1}{2} \sec^2 \frac{x}{2} dx \).
Then \( du = dx \) and \( v = \int \frac{1}{2} \sec^2 \frac{x}{2} dx = \frac{1}{2} \cdot 2 \tan \frac{x}{2} = \tan \frac{x}{2} \).
Using \( \int u dv = uv - \int v du \):
\( \int \frac{1}{2} x \sec^2 \frac{x}{2} dx = x \tan \frac{x}{2} - \int \tan \frac{x}{2} dx \). This is a helpful intermediate step.
Now substitute this back into the original integral for \( I \):
\( I = \left( x \tan \frac{x}{2} - \int \tan \frac{x}{2} dx \right) + \int \tan \frac{x}{2} dx \)
The integral terms cancel out.
\( I = x \tan \frac{x}{2} + C \).
In simple words: We used half-angle formulas to simplify the fraction. Then, we used a method called integration by parts for one part of the integral, which caused other parts to cancel out, leading to a simple answer.

🎯 Exam Tip: When an integrand contains both algebraic (like \( x \)) and trigonometric functions (like \( \sin x \), \( \cos x \)), trigonometric identities often help transform it into a form that can be solved by integration by parts or direct formulas, often leading to cancellations.

Question 16. \( \int e^x\left(\frac{1-x}{1+x^2}\right)^2 d x \)
Answer:
Let \( I = \int e^x \left(\frac{1-x}{1+x^2}\right)^2 d x \)
First, expand and simplify the term inside the integral:
\( \left(\frac{1-x}{1+x^2}\right)^2 = \frac{(1-x)^2}{(1+x^2)^2} = \frac{1 - 2x + x^2}{(1+x^2)^2} \)
We can rewrite the numerator to separate a term that matches the denominator:
\( = \frac{(1+x^2) - 2x}{(1+x^2)^2} \)
Now, separate the terms:
\( = \frac{1+x^2}{(1+x^2)^2} - \frac{2x}{(1+x^2)^2} \)
\( = \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2} \). This manipulation is a standard technique for these types of integrals.
So, the integral becomes:
\( I = \int e^x \left( \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2} \right) dx \)
This is in the form \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \).
Here, let \( f(x) = \frac{1}{1+x^2} \).
Then \( f'(x) = \frac{d}{dx} \left( (1+x^2)^{-1} \right) = -1 (1+x^2)^{-2} \cdot (2x) = -\frac{2x}{(1+x^2)^2} \).
So, the integral is in the form \( \int e^x (f(x) + f'(x)) dx \).
Therefore, \( I = e^x \cdot \frac{1}{1+x^2} + C = \frac{e^x}{1+x^2} + C \).
In simple words: We expanded the square and rearranged the terms inside the integral. This helped us find a function and its derivative, which let us use a special rule to solve the integral.

🎯 Exam Tip: When you see an integrand with \( e^x \) multiplied by a rational function raised to a power, try algebraic manipulation to express it in the \( f(x) + f'(x) \) form. Look for patterns like \( \frac{1}{g(x)} \) and \( -\frac{g'(x)}{(g(x))^2} \).