Indefinite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{Let }}I = \int {\left( {1 + x - \frac{1}{x}} \right){e^{x\,\, + \,\,\frac{1}{x}}}dx} \cr & = \int {{e^{x\,\, + \,\,\frac{1}{x}}}} dx + \int {\left( {x - \frac{1}{x}} \right){e^{x\,\, + \,\,\frac{1}{x}}}dx} \cr & = x.{e^{x\,\, + \,\,\frac{1}{x}}} - \int {x\left( {1 - \frac{1}{{{x^2}}}} \right){e^{x\,\, + \,\,\frac{1}{x}}}dx + \int {\left( {x - \frac{1}{x}} \right){e^{x\,\, + \,\,\frac{1}{x}}}dx} } \cr & = x.{e^{x\,\, + \,\,\frac{1}{x}}} - \int {\left( {x - \frac{1}{{{x}}}} \right){e^{x\,\, + \,\,\frac{1}{x}}}dx + \int {\left( {x - \frac{1}{x}} \right){e^{x\,\, + \,\,\frac{1}{x}}}dx} } \cr & = x.{e^{x\,\, + \,\,\frac{1}{x}}} + C \cr} $$

$$\eqalign{ & I = \int {\frac{{dx}}{{x\left( {{x^n} + 1} \right)}}} = \int {\frac{{dx}}{{{x^{n + 1}}\left( {1 + \frac{1}{{{x^n}}}} \right)}}} \cr & {\text{Put }}1 + \frac{1}{{{x^n}}} = t \Rightarrow - \frac{n}{{{x^{n + 1}}}}dx = dt \cr & I = - \frac{1}{n}\int {\frac{{dt}}{t}} \cr & \Rightarrow I = - \frac{1}{n}\ln \,t + C \cr & \Rightarrow I = - \frac{1}{n}\ln \left( {1 + \frac{1}{{{x^n}}}} \right) + C \cr & \Rightarrow I = - \frac{1}{n}\ln \left( {\frac{{{x^n} + 1}}{{{x^n}}}} \right) + C \cr & \therefore \,A = - \frac{1}{n} \cr} $$

$$\eqalign{ & {\text{Let }}I = \int {\frac{{\log \,x}}{{{{\left( {1 + \log \,x} \right)}^2}}}dx} \cr & {\text{Put }}\log \,x = t \Rightarrow \frac{1}{x}dx = dt \cr & I = \int {\frac{{{e^t}t}}{{{{\left( {1 + t} \right)}^2}}}dt} \cr & = \int {\frac{{{e^t}.\left( {t + 1 - 1} \right)}}{{{{\left( {1 + t} \right)}^2}}}} dt \cr & = \int {\frac{{{e^t}\left( {1 + t} \right)}}{{{{\left( {1 + t} \right)}^2}}}} dt - \int {\frac{{{e^t}}}{{{{\left( {1 + t} \right)}^2}}}dt} \cr & = \int {\frac{{{e^t}}}{{1 + t}}dt - } \int {\frac{{{e^t}}}{{{{\left( {1 + t} \right)}^2}}}dt} \cr & = \frac{{{e^t}}}{{1 + t}} - \int { - {e^t}\frac{1}{{{{\left( {1 + t} \right)}^2}}}dt} - \int {\frac{{{e^t}}}{{{{\left( {1 + t} \right)}^2}}}dt} \cr & = \frac{{{e^t}}}{{1 + t}} + \int {^{{e^t}}\frac{1}{{{{\left( {1 + t} \right)}^2}}}dt} - \int {\frac{{{e^t}}}{{{{\left( {1 + t} \right)}^2}}}dt} \cr & = \frac{{{e^t}}}{{1 + t}} + c \cr & = \frac{x}{{1 + \log \,x}} + c \cr} $$

Let $$I = \int {{f^{ - 1}}\left( x \right)dx} $$     and $${f^{ - 1}}\left( x \right) = t \Rightarrow x = f\left( t \right) \Rightarrow dx = f'\left( t \right)\,dt$$
Put value of $$dx$$  and $${f^{ - 1}}\left( x \right)$$   in $$I,$$ we get $$I = \int {tf'\left( t \right)dt} $$
Now, integrate it by parts, $$I = tf\left( t \right) - \int {f\left( t \right)} dt$$
Given, $$\int {f\left( x \right)dx = g} \left( x \right) + C$$
$$\therefore \,I = tf\left( t \right) - \left[ {g\left( t \right)} \right] + C$$
Now, by putting value of $$t,\,f\left( t \right)$$   and $$g\left( t \right)$$  we get,
$$I = x{f^{ - 1}}\left( x \right)g\left[ {{f^{ - 1}}\left( x \right)} \right] + C$$

$$\frac{d}{{dx}}\left( {{x^x}} \right) = x.{x^{x - 1}} + {x^x}.\log \,x = {x^x}\left( {1 + \log \,x} \right)$$

$$\eqalign{ & {\text{Let }}I = \int {\frac{{dx}}{{{x^{29}}\left( {1 - \frac{6}{{{x^7}}}} \right)}}} \cr & {\text{Put }}1 - \frac{6}{{{x^7}}} = p \Rightarrow \frac{{42}}{{{x^8}}}dx = dp{\text{ and }}{x^7} = \frac{6}{{1 - p}} \cr & \therefore \,I = \frac{1}{{42}}\int {\frac{{{{\left( {1 - p} \right)}^3}}}{{{{\left( 6 \right)}^3}p}}dp} \cr & = \frac{1}{{\left( {42} \right)\left( {216} \right)}}\int {\frac{{1 - {p^3} - 3p + 3{p^2}}}{p}} dp \cr & = \frac{1}{{9072}}\int {\left( {\frac{1}{p} - {p^2} - 3 + 3p} \right)} dp \cr & = \frac{1}{{9072}}\left( {\log \,p - \frac{{{p^3}}}{3} - 3p + \frac{3}{2}{p^2}} \right) + c \cr & = \frac{1}{{54432}}\left( {6\,\ln \,p - 2{p^3} - 18p + 9{p^2}} \right) + c \cr & = \frac{1}{{54432}}\left( {\ln \,{p^6} + 9{p^2} - 2{p^3} - 18p} \right) + c \cr & A = \frac{1}{{54432}},\,\,p = \left( {\frac{{{x^7} - 6}}{{{x^7}}}} \right) \cr} $$

$$\eqalign{ & \int {x\,\log \left( {1 + \frac{1}{x}} \right)dx} \cr & = \log \left( {1 + \frac{1}{x}} \right).\frac{{{x^2}}}{2} - \int {\frac{x}{{x + 1}}.\left( { - \frac{1}{{{x^2}}}} \right).\frac{{{x^2}}}{2}} dx \cr & = \frac{{{x^2}}}{2}\log \left( {\frac{{x + 1}}{x}} \right).\frac{{{x^2}}}{2} + \frac{1}{2}\int {\frac{{x + 1 - 1}}{{x + 1}}} dx \cr & = \frac{{{x^2}}}{2}\log \left( {\frac{{x + 1}}{x}} \right) + \frac{1}{2}x - \frac{1}{2}\log \left( {x + 1} \right) + c \cr & = \left( {\frac{{{x^2} - 1}}{2}} \right)\log \left( {x + 1} \right) - \frac{{{x^2}}}{2}\log \,x + \frac{1}{2}x + c \cr} $$

$$\eqalign{ & I = \int {\frac{{dx}}{{2\cos \left( {x - \frac{\pi }{3}} \right)}}} \cr & \,\,\,\,\, = \frac{1}{2}\int {\sec \left( {x - \frac{\pi }{3}} \right)dx} \cr & \,\,\,\,\, = \frac{1}{2}\log \left\{ {\sec \left( {x - \frac{\pi }{3}} \right) + \tan \left( {x - \frac{\pi }{3}} \right)} \right\} + k \cr & \,\,\,\,\, = \frac{1}{2}\log \,\tan \left( {\frac{\pi }{2} + \frac{{x - \frac{\pi }{3}}}{2}} \right) + k \cr & \,\,\,\,\, = \frac{1}{2}\log \,\tan \left( {\frac{x}{2} + \frac{\pi }{3}} \right) + k \cr} $$

$$\eqalign{ & \int {g\left( x \right)\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} \cr & = \int {g\left( x \right)f\left( x \right)dx} + \int {g\left( x \right)f'\left( x \right)dx} \cr & = f\left( x \right)\int {g\left( x \right)dx - } \int {\left\{ {f'\left( x \right)\int {g\left( x \right)} dx} \right\}dx + \int {g\left( x \right)f'\left( x \right)dx} } \cr & = f\left( x \right)g\left( x \right) - \int {f'\left( x \right)g\left( x \right)dx + \int {g\left( x \right)f'\left( x \right)dx} } \,\,\,\left[ {{\text{Given }}\int {g\left( x \right)dx = g\left( x \right)} } \right] \cr & = f\left( x \right)g\left( x \right) + c \cr} $$

$$\eqalign{ & {\text{Let }}I = \int {{{\sec }^n}x\,\tan \,x\,dx} \cr & {\text{Put, }}\sec \,x = t \Rightarrow \sec \,x\,\tan \,x\,dx = dt \cr & \therefore \,I = \int {{t^n}.\frac{{dt}}{t}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \int {{t^{n - 1}}dt} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{t^n}}}{n} + c \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\sec }^n}x}}{n} + c \cr} $$
where $$'c’$$ is a constant of integration.