Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Logarithmic integration. Click on this to access more such competitive quizzes on Integral Calculus.

Logarithmic integration Quiz 2

mathematics

 Integrate the function$\dfrac{y}{x}=\log \left( 78{{x}^{2}}-123.099 \right)$ with respect to x.

 First let $v=78{{x}^{2}}-123.099$ then substitute its value and integrate using the formula, $\int{\log udu}=\log u\int{du}-\int{\left( \dfrac{d}{du}\log u\int{du} \right)du} $ where $c$ is a constant.

Given equation can be written as, $y=x\log \left( 78{{x}^{2}}-123.099 \right) \\ $ Let $v=78{{x}^{2}}-123.099\Rightarrow dv=156xdx \\  \Rightarrow \int{ydx=\dfrac{1}{156}\int{\log vdv}} \\ $ On Integrating by part, $ \int{ydx}=\dfrac{v}{156}(\log v-1)+c=\dfrac{78{{x}^{2}}-123.099}{156}\left[ \log \left( 78{{x}^{2}}-123.099 \right)-1 \right]+c$

 Find the antiderivative of ${y{{e}^{-7x}}}=\log \left( {{e}^{7x}}+999 \right)$ with respect to $x$.

 First let ${{e}^{7x}}+999$ then substitute its value and integrate using the formula, $\int{\log udu}=\dfrac{1}{u}+c$ and $ \int{{{e}^{x}}dx={{e}^{x}}+c}$ where $c$ is a constant.

Given equation can be written as, $y={{e}^{7x}}\log \left( {{e}^{7x}}+999 \right) \\ $ Let, $v={{e}^{7x}}+999 \Rightarrow dv=7{{e}^{7x}}dx \\  \therefore \int{ydv=}\dfrac{1}{7}\int{\log vdv=}\dfrac{1}{7}I' \\ \Rightarrow I'=u\log u-\int{\left( \dfrac{1}{u}\times u \right)du}=u\log u-\int{du}=u(\log u-1)+c' \\  \therefore \int{ydv=}\dfrac{1}{7}v(\log v-1)+c'=\dfrac{{{e}^{7x}}+999}{7}\left[ \log \left( {{e}^{7x}}+999 \right)-1 \right]+c$

 Evaluate: $I=\int{(5a{{\theta }^{4}}+b\theta )\log \left( 2a{{\theta }^{5}}+b{{\theta }^{2}}+c \right)d\theta }$.

 First let $x=2a{{\theta }^{5}}+b{{\theta }^{2}}+c$ then substitute its value and integrate using the formula, $\int{\log udu}=\dfrac{1}{u}+c$, where $c$ is a constant.

$  I=\int{(5a{{\theta }^{4}}+b\theta )\log \left( 2a{{\theta }^{5}}+b{{\theta }^{2}}+c \right)d\theta } \\  $ Let $x=2a{{\theta }^{5}}+b{{\theta }^{2}}+c\Rightarrow dx=\left( 10a{{\theta }^{4}}+2b\theta  \right)d\theta  \\  \Rightarrow I=\dfrac{1}{2}\int{\log xdx=\dfrac{1}{2}I'} \\  I'=\int{\log xdx}=\log x\int{dx}-\int{\left( \dfrac{d}{dx}\log x\int{dx} \right)dx} \\  \Rightarrow I'=x\log x-\int{\left( \dfrac{1}{x}\times x \right)dx}=x\log x-\int{dx}=x(\log x-1)+c' \\  \Rightarrow I=\dfrac{2a{{\theta }^{5}}+b{{\theta }^{2}}+c}{2}\left[ \log \left( 2a{{\theta }^{5}}+b{{\theta }^{2}}+c \right)-1 \right]+c$

 Evaluate: $\int{\cos x\sqrt{\text{cosec} x}\log \left( \sqrt{\text{cosec} x} \right)dx}$.

 First let $u=\text{cosec} x$ then substitute its value and integrate using the formula, $\int{\log udu}=\dfrac{1}{u}+c$, where $c$ is a constant.

$\int{\cos x\sqrt{\text{cosec} x}\log \left( \sqrt{\text{cosec} x} \right)dx} \\ $ Let, $u=\text{cosec} x \\ I=\int \:-\dfrac{\ln \left(u\right)}{2u\sqrt{u}}du \\ \text{Apply integration by parts where,} u=\ln \left(u\right)\text{ and }v’=\dfrac{1}{u\sqrt{u}} \\ I=-\dfrac{1}{2}\left(-\dfrac{2\ln \left(u\right)}{\sqrt{u}}-\int \:-\dfrac{2}{u\sqrt{u}}du\right) \\ I=-\dfrac{1}{2}\left(-\dfrac{2\ln \left(u\right)}{\sqrt{u}}-\dfrac{4}{\sqrt{u}}\right)\\ I=-\dfrac{1}{2}\left(-\dfrac{2\ln \left(\csc \left(x\right)\right)}{\sqrt{\csc \left(x\right)}}-\dfrac{4}{\sqrt{\csc \left(x\right)}}\right) \\ I=\sqrt{\sin \left(x\right)}\left(\ln \left(\csc \left(x\right)\right)+2\right)+C$

 Integrate the function $\dfrac{y+{{e}^{2x}}}{{{x}^{4}}}={{\left( \log x \right)}^{2}}$ with respect to $x$ and determine which of the following options is true?

 First integrate using the by-parts formula, $\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}dx$, then use $\int{\log udu}=\dfrac{1}{u}+c$, where $c$ is a constant.

$\dfrac{y+{{e}^{2x}}}{{{x}^{4}}}={{\left( \log x \right)}^{2}} \\ \int \:x^4\left(\ln \left(x\right)\right)^2-e^{2x}dx \\ =\int \:x^4\left(\ln \left(x\right)\right)^2dx-\int \:e^{2x}dx \\ =\dfrac{1}{5}x^5\ln ^2\left(x\right)-\int \dfrac{2}{5}x^4\ln \left(x\right)dx-\dfrac{1}{2}e^{2x} \\ =\dfrac{1}{5}x^5\ln ^2\left(x\right)-\dfrac{2}{5}\left(\dfrac{1}{5}x^5\ln \left(x\right)-\dfrac{x^5}{25}\right)-\dfrac{1}{2}e^{2x} + C$