Question

intregration of log tan inverse x divided by x square + 1​

Answer 1

Topic :-

Indefinite Integration

To Integrate :-

$\displaystyle \int \dfrac{\log(\arctan x)}{1+x^2}dx$

Solution :-

$\displaystyle \int \dfrac{\log(\arctan x)}{1+x^2}dx$

$Substitute\:\arctan x = t.$

$Upon\:differentiation,$

$\dfrac{1}{1+x^2}\:dx =dt$

$\displaystyle \int \log(\arctan x)\cdot\left(\dfrac{1}{1+x^2}\cdot dx\right)$

$\displaystyle \int \log(t)\:dt$

$It\:can\:be\:written\:as,$

$\displaystyle \int t^0\cdot\log(t)\:dt$

$(\because t^0=1)$

$\log t\displaystyle\int t^0\:dt-\int \left[\dfrac{d(\log t)}{dt}\cdot \int t^0\:dt\right]dt$

$\left ( \because \displaystyle \int u\cdot v\:dx=u\int v \:dx-\int\left[ \dfrac{du}{dx}\int v\:dx\right]dx \right)$

$\log t\cdot (t)-\displaystyle \int \left[\dfrac{d(\log t)}{dt}\cdot (t)\right]dt$

$\left(\because \displaystyle \int x^n\:dx=\dfrac{x^{n+1}}{n+1}+C\right)$

$t\cdot \log t-\displaystyle \int \left[\dfrac{1}{t}\cdot t\right]dt$

$\left(\because \dfrac{d(\log x)}{dx}=\dfrac{1}{x}\right)$

$t\cdot \log t-\displaystyle \int 1\:dt$

$t\cdot \log t-\displaystyle \int t^0\:dt$

$(\because t^0=1)$

$t\cdot \log t-t+C$

$\left(\because \displaystyle \int x^n\:dx=\dfrac{x^{n+1}}{n+1}+C\right)$

$Put\:back\:t=\arctan x,$

$(\arctan x)\cdot \log (\arctan x)-(\arctan x)+C$

$or$

$(\arctan x)(\log (\arctan x)-1)+C$

Answer :-

$\underline{\boxed{(\arctan x)(\log (\arctan x)-1)+C}}$

$\bold{Note :}\:\arctan x=\tan^{-1}x$