JEE Mathematics Indefinite Integration MCQs Set 10

Answer: (a) \( \frac{23 x}{41} + \frac{2 \log{\left(4 \sin{\left(x \right)} + 5 \cos{\left(x \right)} \right)}}{41}+c \)

Explanation:
1. Write the numerator as \( A(4\sin x+5\cos x)+B(4\cos x-5\sin x) \): \( 4A-5B=2 \), \( 5A+4B=3 \), so \( A=\frac{23}{41} \), \( B=\frac2{41} \).
2. The integral is \( \frac{23}{41}x+\frac{2}{41}\log|4\sin x+5\cos x|+c \).

Teacher's Note:
1. Express the numerator as a multiple of the denominator plus a multiple of its derivative.
2. Solve the two linear equations for A and B.

Answer: (a) \( \frac{\sqrt{2} \log{\left(\tan{\left(\frac{x}{2} + \frac{3 \pi}{8} \right)} \right)}}{2}+c \)

Explanation:
1. \( \cos x-\sin x=\sqrt2\cos\left(x+\frac\pi4\right) \).
2. \( \frac1{\sqrt2}\int\sec\left(x+\frac\pi4\right)dx=\frac{1}{\sqrt2}\log\tan\left(\frac\pi4+\frac x2+\frac\pi8\right)=\frac1{\sqrt2}\log\tan\left(\frac x2+\frac{3\pi}8\right)+c \).

Teacher's Note:
1. Combine cos and sin into a single cosine.
2. The integral of sec u is log tan (pi/4 + u/2).

Answer: (a) \( \frac{12 x}{13} - \frac{5 \log{\left(2 \sin{\left(x \right)} + 3 \cos{\left(x \right)} \right)}}{13}+c \)

Explanation:
1. Write \( 3\sin x+2\cos x=A(3\cos x+2\sin x)+B(-3\sin x+2\cos x) \): \( 2A-3B=3 \), \( 3A+2B=2 \), so \( A=\frac{12}{13} \), \( B=-\frac5{13} \).
2. The integral is \( \frac{12}{13}x-\frac5{13}\log(3\cos x+2\sin x)+c \).

Teacher's Note:
1. Use the same numerator splitting method.
2. Check the sign of B.

Answer: (a) \( 6 \sqrt{x^{2} - 9 x + 20} + 34 \log{\left(x + \sqrt{x^{2} - 9 x + 20} - \frac{9}{2} \right)}+c \)

Explanation:
1. \( (x-5)(x-4)=x^2-9x+20 \) and \( 6x+7=3(2x-9)+34 \).
2. \( 3\int\frac{2x-9}{\sqrt{x^2-9x+20}}dx=6\sqrt{x^2-9x+20} \) and \( 34\int\frac{dx}{\sqrt{\left(x-\frac92\right)^2-\frac14}}=34\log\left|x-\frac92+\sqrt{x^2-9x+20}\right| \).
3. The sum is \( 6\sqrt{x^2-9x+20}+34\log\left|x-\frac92+\sqrt{x^2-9x+20}\right|+c \).

Teacher's Note:
1. Split the numerator into a multiple of the derivative of the quadratic plus a constant.
2. The second part gives a logarithm of the standard form.

Answer: (b) \( \frac{\sqrt{2} \tan^{-1}{\left(\frac{\sqrt{2} \left(x^{2} - 1\right)}{2 x} \right)}}{2}+c \)

Explanation:
1. Divide numerator and denominator by \( x^2 \): \( \frac{1+\frac1{x^2}}{x^2+\frac1{x^2}}=\frac{1+\frac1{x^2}}{\left(x-\frac1x\right)^2+2} \).
2. Put \( t=x-\frac1x \), \( dt=\left(1+\frac1{x^2}\right)dx \): \( \int\frac{dt}{t^2+2}=\frac1{\sqrt2}\tan^{-1}\frac{t}{\sqrt2}=\frac1{\sqrt2}\tan^{-1}\frac{x^2-1}{\sqrt2x}+c \).

Teacher's Note:
1. Divide by x squared and complete the square in x minus 1 over x.
2. Use the arctangent standard form.

Answer: (b) \( \frac{\sqrt{2} \tan^{-1}{\left(\frac{\sqrt{2} x}{\sqrt{1 - x^{2}}} \right)}}{2}+c \)

Explanation:
1. Put \( t=\frac{x}{\sqrt{1-x^2}} \). Then \( dt=\frac{dx}{(1-x^2)^{3/2}} \) and \( 1+x^2=\frac{1+2t^2}{1+t^2} \), \( 1-x^2=\frac1{1+t^2} \).
2. The integral becomes \( \int\frac{dt}{1+2t^2}=\frac1{\sqrt2}\tan^{-1}(\sqrt2t)=\frac1{\sqrt2}\tan^{-1}\frac{\sqrt2x}{\sqrt{1-x^2}}+c \).

Teacher's Note:
1. The substitution t = x over root of 1 - x squared simplifies both factors.
2. Express x squared through t squared.

Answer: (a) \( \log{\left(\frac{\left(x - 2\right)^{2}}{x - 1} \right)}+c \)

Explanation:
1. \( \frac{x}{(x-2)(x-1)}=\frac{2}{x-2}-\frac{1}{x-1} \).
2. \( \int=2\log_e(x-2)-\log_e(x-1)+p=\log_e\frac{(x-2)^2}{x-1}+p \).

Teacher's Note:
1. Use the cover up rule for partial fractions.
2. Combine the logs with the power rule.

Answer: (c) \( \log{\left(\frac{x + 1}{x + 2} \right)}+c \)

Explanation:
1. \( \frac{1}{(x+1)(x+2)}=\frac1{x+1}-\frac1{x+2} \).
2. \( \int=\log(x+1)-\log(x+2)=\log\frac{x+1}{x+2}+c \).

Teacher's Note:
1. Write 1 as (x + 2) minus (x + 1) in the numerator.
2. The difference of logs is the log of a quotient.

Answer: (c) \( \frac{\log{\left(\frac{x + 1}{1 - x} \right)}}{2}+c \)

Explanation:
1. \( \frac{1}{1-x^2}=\frac12\left[\frac{1}{1+x}+\frac{1}{1-x}\right] \).
2. \( \int=\frac12\left[\log|1+x|-\log|1-x|\right]=\frac12\log\left|\frac{1+x}{1-x}\right|+c \).

Teacher's Note:
1. Use partial fractions or the formula for 1 over a squared minus x squared.
2. Watch the sign: the integral of 1 over 1 - x is minus log of 1 - x.

Answer: (b) \( \log{\left(\left(x - 3\right)^{2} \right)} - \log{\left(x - 2 \right)}+c \)

Explanation:
1. \( \frac{x-1}{(x-3)(x-2)}=\frac{2}{x-3}-\frac{1}{x-2} \).
2. \( \int=2\log(x-3)-\log(x-2)=\log(x-3)^2-\log(x-2)+c \).

Teacher's Note:
1. Find the constants by the cover up rule.
2. Write 2 log as the log of a square.

Answer: (d) \( A=\dfrac25,\ B=\dfrac{-2}{5\sqrt{15}},\ f(x)=\dfrac{4\tan\frac x2+1}{\sqrt{15}} \)

Explanation:
1. \( \frac{1}{(\sin x+4)(\sin x-1)}=\frac15\left[\frac{1}{\sin x-1}-\frac{1}{\sin x+4}\right] \). Put \( t=\tan\frac x2 \).
2. \( \int\frac{dx}{\sin x-1}=-2\int\frac{dt}{(t-1)^2}=\frac{2}{t-1} \) and \( \int\frac{dx}{\sin x+4}=\int\frac{dt}{2t^2+t+2}=\frac{2}{\sqrt{15}}\tan^{-1}\frac{4t+1}{\sqrt{15}} \).
3. So the integral is \( \frac25\cdot\frac{1}{\tan\frac x2-1}-\frac{2}{5\sqrt{15}}\tan^{-1}\frac{4\tan\frac x2+1}{\sqrt{15}}+c \).

Teacher's Note:
1. Split with partial fractions in sin x first.
2. Then use the half angle substitution on each part.

Answer: (a) \( \log{\left(\frac{x - 2}{x - 1} \right)} + \frac{1}{x - 1}+c \)

Explanation:
1. \( \frac{1}{(x-1)^2(x-2)}=\frac{-1}{x-1}-\frac{1}{(x-1)^2}+\frac{1}{x-2} \).
2. \( \int=-\log|x-1|+\frac{1}{x-1}+\log|x-2|=\log\left|\frac{x-2}{x-1}\right|+\frac1{x-1}+c \).

Teacher's Note:
1. Write the partial fractions with a repeated factor.
2. Integrate the squared term with the power rule.

Answer: (a) \( \log{\left(x + 1 \right)} + \frac{2}{x + 1}+c \)

Explanation:
1. \( \frac{x-1}{(x+1)^2}=\frac{(x+1)-2}{(x+1)^2}=\frac1{x+1}-\frac{2}{(x+1)^2} \).
2. \( \int=\log(x+1)+\frac{2}{x+1}+c \).

Teacher's Note:
1. Write the numerator in terms of x + 1.
2. Integrating minus 2 over a square gives plus 2 over the base.

Answer: (a) \( \log{\left(x \right)} - \log{\left(1 - x \right)}+c \)

Explanation:
1. \( \frac{1}{x(1-x)}=\frac1x+\frac1{1-x} \).
2. \( \int=\log x-\log(1-x)+c \).

Teacher's Note:
1. Use partial fractions.
2. The integral of 1 over 1 - x carries a minus sign.

Answer: (b) \( \frac{a \log{\left(\frac{- a + x}{a + x} \right)}}{2} + x+c \)

Explanation:
1. \( \frac{x^2}{x^2-a^2}=1+\frac{a^2}{x^2-a^2} \).
2. \( \int=x+a^2\cdot\frac1{2a}\log\frac{x-a}{x+a}=x+\frac a2\log\frac{x-a}{x+a}+c \).

Teacher's Note:
1. Divide first because the degrees are equal.
2. Use the standard 1 over x squared minus a squared formula.

Answer: (c) \( - \frac{\tan^{-1}{\left(\frac{x}{2} \right)}}{6} + \frac{\tan^{-1}{\left(x \right)}}{3}+c \)

Explanation:
1. \( \frac{1}{(x^2+1)(x^2+4)}=\frac13\left[\frac{1}{x^2+1}-\frac{1}{x^2+4}\right] \).
2. \( \int=\frac13\tan^{-1}x-\frac13\cdot\frac12\tan^{-1}\frac x2=\frac13\tan^{-1}x-\frac16\tan^{-1}\frac x2+c \).

Teacher's Note:
1. Treat x squared as a single variable for partial fractions.
2. The second integral brings a factor 1 over 2.

Answer: (b) \( - \sqrt{2} \tan^{-1}{\left(\frac{\sqrt{2} x}{2} \right)} + \sqrt{3} \tan^{-1}{\left(\frac{\sqrt{3} x}{3} \right)}+c \)

Explanation:
1. With \( u=x^2 \): \( \frac{u}{(u+2)(u+3)}=\frac{-2}{u+2}+\frac{3}{u+3} \).
2. \( \int=-2\cdot\frac{1}{\sqrt2}\tan^{-1}\frac x{\sqrt2}+3\cdot\frac1{\sqrt3}\tan^{-1}\frac x{\sqrt3}=-\sqrt2\tan^{-1}\frac{x}{\sqrt2}+\sqrt3\tan^{-1}\frac{x}{\sqrt3}+c \).

Teacher's Note:
1. Use the cover up rule on x squared as the variable.
2. Each term gives an arctangent.

Answer: (c) \( \frac{\log{\left(\frac{- a^{2} + x^{2}}{- b^{2} + x^{2}} \right)}}{2 a^{2} - 2 b^{2}}+c \)

Explanation:
1. Put \( u=x^2 \), \( x\,dx=\frac{du}{2} \): \( \frac12\int\frac{du}{(u-a^2)(u-b^2)} \).
2. \( \frac{1}{(u-a^2)(u-b^2)}=\frac{1}{a^2-b^2}\left[\frac1{u-a^2}-\frac1{u-b^2}\right] \), so the result is \( \frac{1}{2(a^2-b^2)}\log\frac{x^2-a^2}{x^2-b^2}+c \).

Teacher's Note:
1. Substitute u = x squared to get a rational function.
2. Do not forget the factor 1 over 2.

Answer: (a) \( \dfrac12,\ \dfrac34 \)

Explanation:
1. With \( u=x^2 \): \( \frac{2u+1}{(u-4)(u-1)}=\frac{3}{u-4}-\frac{1}{u-1} \).
2. \( \int\frac{3\,dx}{x^2-4}=\frac34\log\frac{x-2}{x+2} \) and \( -\int\frac{dx}{x^2-1}=\frac12\log\frac{x+1}{x-1} \).
3. Hence \( a=\frac12 \) and \( b=\frac34 \).

Teacher's Note:
1. Partial fractions in x squared work as in earlier questions.
2. Match the two logarithmic terms with the given form.

Answer: (d) \( \left(\dfrac12,\dfrac12\right) \)

Explanation:
1. With \( u=x^2 \): \( \frac{2u+3}{(u-1)(u+4)}=\frac{1}{u-1}+\frac{1}{u+4} \).
2. \( \int\frac{dx}{x^2-1}=\frac12\log\frac{x-1}{x+1} \) and \( \int\frac{dx}{x^2+4}=\frac12\tan^{-1}\frac x2 \).
3. So \( a=\frac12 \) and \( b=\frac12 \).

Teacher's Note:
1. Use the cover up rule with x squared as the variable.
2. Each part is a standard integral.

Answer: (b) \( \frac{\log{\left(\frac{x^{4} - 1}{x^{4}} \right)}}{4}+c \)

Explanation:
1. Multiply numerator and denominator by \( x^3 \): \( \frac{x^3}{x^4(x^4-1)} \). Put \( u=x^4 \), \( x^3dx=\frac{du}{4} \).
2. \( \frac14\int\frac{du}{u(u-1)}=\frac14\log\frac{u-1}{u}=\frac14\log\frac{x^4-1}{x^4}+k \).

Teacher's Note:
1. Multiply by x cubed to make x to the fourth the natural variable.
2. Then partial fractions in u.

Answer: (c) \( \frac{\log{\left(e^{x} - 1 \right)}}{3} - \frac{\log{\left(e^{x} + 2 \right)}}{3}+c \)

Explanation:
1. \( e^x+1-2e^{-x}=\frac{e^{2x}+e^x-2}{e^x}=\frac{(e^x-1)(e^x+2)}{e^x} \).
2. Put \( t=e^x \): \( \int\frac{dt}{(t-1)(t+2)}=\frac13\log\frac{t-1}{t+2}=\frac13\log(e^x-1)-\frac13\log(e^x+2)+c \).

Teacher's Note:
1. Multiply numerator and denominator by e to the x.
2. Factorise the quadratic in e to the x.

Answer: (b) \( x + \frac{7 \sqrt{3} \log{\left(\frac{x - 2 \sqrt{3} + 3}{x + 3 + 2 \sqrt{3}} \right)}}{4} - 3 \log{\left(x^{2} + 6 x - 3 \right)}+c \)

Explanation:
1. \( \frac{x^2}{x^2+6x-3}=1+\frac{-6x+3}{x^2+6x-3}=1-\frac{3(2x+6)}{x^2+6x-3}+\frac{21}{x^2+6x-3} \).
2. With \( x^2+6x-3=(x+3)^2-(2\sqrt3)^2 \), \( \int\frac{21\,dx}{(x+3)^2-12}=\frac{21}{4\sqrt3}\log\left|\frac{x+3-2\sqrt3}{x+3+2\sqrt3}\right| \).
3. The integral is \( x-3\log|x^2+6x-3|+\frac{21}{4\sqrt3}\log\left|\frac{x+3-2\sqrt3}{x+3+2\sqrt3}\right|+c \).

Teacher's Note:
1. Divide first because the degrees are equal.
2. Write the new numerator as a multiple of the derivative of the denominator plus a constant.

Answer: (b) \( \log{\left(\frac{e^{x} + 1}{e^{x} + 2} \right)}+c \)

Explanation:
1. Put \( t=e^x \): \( \int\frac{dt}{(1+t)(2+t)}=\int\left[\frac1{1+t}-\frac1{2+t}\right]dt \).
2. \( =\log\frac{1+t}{2+t}=\log\frac{1+e^x}{2+e^x}+c \).

Teacher's Note:
1. Substitute t = e to the x so that e to the x dx becomes dt.
2. Use partial fractions.

Answer: (d) \( - \frac{x^{2}}{2} + \log{\left(\frac{x - 1}{x + 1} \right)}+c \)

Explanation:
1. \( x^3-x-2=-x(1-x^2)-2 \), so the integrand is \( -x-\frac{2}{1-x^2} \).
2. \( \int=-\frac{x^2}{2}-2\cdot\frac12\log\left|\frac{1+x}{1-x}\right|=\log\frac{x-1}{x+1}-\frac{x^2}{2}+c \).

Teacher's Note:
1. Divide the numerator by 1 - x squared.
2. Use the formula for 1 over 1 - x squared with correct signs.