Question

integration of tan inverse root x

Answer 1

integral tan^-1√x.1= now use integral by parts. I can't solve now but it may help u

Answer 2

Answer:

The integration of $\int{\arctan \sqrt{x}} \, dx$ is $(x+1)\arctan \sqrt{x}-\sqrt{x}$

Step-by-step explanation:

We have to integrate $L=\int{\arctan \sqrt{x}} \, dx$

Let, $u=\sqrt{x}$

Then, $du= \frac{1}{2\sqrt{x}}dx$ i.e. $2u du=dx$

Substituting the values, we get,

$L=\int{2u\arctan u} \, du$

i.e. $L=2\int{u\arctan u} \, du$

Using integration by parts, we have,

$L=2[(\arctan u\int\limits{u} \, du)- (\frac{d}{du}(\arctan u)\int\limits {u} \, du)]$

i.e. $L=2[(\frac{u^2}{2}\arctan u)-(\frac{1}{2}\int\limits {\frac{u^2}{1+u^2} \, du)]$

i.e. $L=2[(\frac{u^2}{2}\arctan u)-(\frac{1}{2}\int\limits {\frac{1+u^2-1}{1+u^2} \, du)]$

i.e. $L=2[(\frac{u^2}{2}\arctan u)-(\frac{1}{2}\int\limits \,du)+(\frac{1}{2}\int\limits {\frac{1}{1+u^2} \, du)$

i.e. $L=2[(\frac{u^2}{2}\arctan u)-(\frac{u}{2})+(\frac{1}{2}\arctan u)]$

i.e. $L=u^2\arctan u-u+\arctan u$

i.e. $L=(u^2+1)\arctan u-u$

i.e. $L=((\sqrt{x})^2+1)\arctan \sqrt{x}-\sqrt{x}$

i.e. $L=(x+1)\arctan \sqrt{x}-\sqrt{x}$

Thus, the integration of $\int{\arctan \sqrt{x}} \, dx$ is $(x+1)\arctan \sqrt{x}-\sqrt{x}$