Question

Evaluate the integral:

$\displaystyle \int \frac{ \int \limits_0^x \tan^{ -1}t \: dt}{ {x}^{3} } dx$
​​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\tt{I=\int\dfrac{\int\limits^{x}_{0}\,tan^{-1}(t)\,dt}{x^3}\,dx}$

Solving the definite integral first,

$\displaystyle\tt{\int^{x}_{0}\,tan^{-1}(t)\,dt}$

Applying Integration By Parts,

$\displaystyle\tt{=\left[tan^{-1}(t)\,\int\,dt\right]^{x}_{0}-\int^{x}_{0}\left\{\dfrac{d}{dt}\,tan^{-1}(t)\cdot\int\,dt\right\}\,dt}$

$\displaystyle\tt{=\left[t\cdot\,tan^{-1}(t)\right]^{x}_{0}-\int^{x}_{0}\left\{\dfrac{1}{1+t^2}\cdot\,t\right\}\,dt}$

$\displaystyle\tt{=\left[t\cdot\,tan^{-1}(t)\right]^{x}_{0}-\dfrac{1}{2}\int^{x}_{0}\dfrac{2t}{1+t^2}\,dt}$

$\displaystyle\tt{=\left[t\cdot\,tan^{-1}(t)\right]^{x}_{0}-\dfrac{1}{2}\left[\ln\left|1+t^2\right|\right]^{x}_{0}}$

$\displaystyle\tt{=x\cdot\,tan^{-1}(x)-\dfrac{1}{2}\ln\left|1+x^2\right|}$

$\displaystyle\tt{=x\cdot\,tan^{-1}(x)-\ln\sqrt{1+x^2}}$

Now,

$\displaystyle\tt{I=\int\dfrac{x\cdot\,tan^{-1}(x)-\ln\sqrt{1+x^2}}{x^3}\,dx}$

$\displaystyle\tt{\implies\,I=\int\dfrac{x\cdot\,tan^{-1}(x)}{x^3}\,dx-\int\dfrac{\ln\sqrt{1+x^2}}{x^3}\,dx}$

$\displaystyle\tt{\implies\,I=\int\dfrac{tan^{-1}(x)}{x^2}\,dx-\int\dfrac{\ln\sqrt{1+x^2}}{x^3}\,dx}$

$\displaystyle\tt{\implies\,I=I_{1}-I_{2}}$

$\bigstar\mathbf{\,\,\purple{Solving\,\,\,I_{1}:}}$

$\displaystyle\tt{\implies\,I_1=\int\dfrac{tan^{-1}(x)}{x^2}\,dx}$

$\sf{\mapsto\,\,\bf{\green{Put\,\,x=tan(\theta)}}}$

$\sf{\mapsto\,\,\bf{\green{dx=sec^2(\theta)\,d\theta}}}$

So,

$\displaystyle\tt{\implies\,I_1=\int\dfrac{tan^{-1}(tan(\theta))}{tan^2(\theta)}\,sec^2(\theta)\,d\theta}$

$\displaystyle\tt{\implies\,I_1=\int\theta\cdot\,cosec^2(\theta)\,d\theta}$

$\displaystyle\tt{\implies\,I_1=\theta\int\,cosec^2(\theta)\,d\theta\,-\int\left\{\dfrac{d}{d\theta}(\theta)\cdot\int\,cosec^2(\theta)\,d\theta\right\}\,d\theta}$

$\displaystyle\tt{\implies\,I_1=-\theta\,cot(\theta)\,+\int\,cot(\theta)\,d\theta}$

$\displaystyle\tt{\implies\,I_1=-\theta\,cot(\theta)\,+\ln|sin(\theta)|}$

$\displaystyle\tt{\implies\,I_1=-tan^{-1}(x)\cdot\,cot(tan^{-1}(x))\,+\ln|sin(tan^{-1}(x))|}$

$\displaystyle\tt{\implies\,I_1=-tan^{-1}(x)\cdot\,\dfrac{1}{x}\,+\ln\left|\dfrac{x}{\sqrt{1+x^2}}\right|}$

$\displaystyle\tt{\implies\,I_1=-\dfrac{tan^{-1}(x)}{x}\,+\ln\left|\dfrac{x}{\sqrt{1+x^2}}\right|+c_{1}}$

$\bigstar\mathbf{\,\,\purple{Solving\,\,\,I_{2}:}}$

$\displaystyle\tt{\implies\,I_{2}=\int\dfrac{\ln\sqrt{1+x^2}}{x^3}\,dx}$

Solving by parts

$\displaystyle\tt{\implies\,I_{2}=\ln\sqrt{1+x^2}\cdot\int\dfrac{1}{x^3}\,dx\,-\int\left[\dfrac{d}{dx}\left\{\ln\sqrt{1+x^2}\right\}\cdot\int\dfrac{dx}{x^3}\right]dx}$

$\displaystyle\tt{\implies\,I_{2}=-\dfrac{\ln\sqrt{1+x^2}}{2x^2}+\int\left[\dfrac{1}{\sqrt{1+x^2}}\cdot\dfrac{x}{\sqrt{1+x^2}}\cdot\dfrac{1}{2x^2}\right]dx}$

$\displaystyle\tt{\implies\,I_{2}=-\dfrac{\ln\sqrt{1+x^2}}{2x^2}+\dfrac{1}{2}\int\dfrac{1}{x(1+x^2)}\,dx}$

$\displaystyle\tt{\implies\,I_{2}=-\dfrac{\ln\sqrt{1+x^2}}{2x^2}+\dfrac{1}{2}\int\left\{\dfrac{1}{x}-\dfrac{x}{x^2+1}\right\}\,dx}$

$\displaystyle\tt{\implies\,I_{2}=-\dfrac{\ln\sqrt{1+x^2}}{2x^2}+\dfrac{1}{2}\int\dfrac{dx}{x}-\dfrac{1}{2}\int\dfrac{x}{x^2+1}\,dx}$

$\displaystyle\tt{\implies\,I_{2}=-\dfrac{\ln\sqrt{1+x^2}}{2x^2}+\dfrac{1}{2}\int\dfrac{dx}{x}-\dfrac{1}{4}\int\dfrac{2x}{x^2+1}\,dx}$

$\displaystyle\tt{\implies\,I_{2}=-\dfrac{\ln\sqrt{1+x^2}}{2x^2}+\dfrac{1}{2}\,\ln|x|-\dfrac{1}{4}\ln|x^2+1|+C}$

$\displaystyle\tt{\implies\,I_{2}=-\dfrac{\ln\sqrt{1+x^2}}{2x^2}+\dfrac{1}{2}\,\ln|x|-\dfrac{1}{2}\ln\sqrt{x^2+1}+C}$

$\displaystyle\tt{\implies\,I_{2}=-\dfrac{\ln\sqrt{1+x^2}}{2x^2}+\dfrac{1}{2}\,\ln\left|\dfrac{x}{\sqrt{x^2+1}}\right|+c_{2}}$

Now,

$\displaystyle\tt{\implies\,I=I_{1}-I_{2}}$

$\displaystyle\tt{\implies\,I=-\dfrac{tan^{-1}(x)}{x}+\ln\left|\dfrac{x}{\sqrt{x^2+1}}\right|+c_{1}+\dfrac{\ln\sqrt{x^2+1}}{2x^2}-\dfrac{1}{2}\,\ln\left|\dfrac{x}{\sqrt{x^2+1}}\right|-c_{2}}$

Put $\sf{Put\,\,\,c_{1}-c_{2}=C}$

So,

$\displaystyle\tt{\implies\,I=-\dfrac{tan^{-1}(x)}{x}+\dfrac{1}{2}\ln\left|\dfrac{x}{\sqrt{x^2+1}}\right|+\dfrac{\ln\sqrt{x^2+1}}{2x^2}+C}$

Answer 2

Answer:

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