Question

solve this pls


$\implies \red{ \int \: \frac{ \sqrt{x} }{1 + x {}^{2} }dx }$
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Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\tt{\int\dfrac{\sqrt{x}}{1+{x}^{2}}\,dx}$

$\bf{\mapsto\,\,\,\,Put\,\,\,\,x={t}^{2}}$

$\bf{\mapsto\,\,\,\,dx=2t\,dt}$

So,

$\displaystyle\tt{\int\dfrac{t}{1+{t}^{4}}\cdot2t\,dt}$

$\displaystyle\tt{=\int\dfrac{2{t}^{2}}{1+{t}^{4}}\,dt}$

$\displaystyle\tt{=\int\dfrac{{t}^{2}+1+{t}^{2}-1}{1+{t}^{4}}\,dt}$

$\displaystyle\tt{=\int\dfrac{{t}^{2}+1}{1+{t}^{4}}\,dt+\int\dfrac{{t}^{2}-1}{1+{t}^{4}}\,dt}$

Dividing num. and den. by t²,

$\displaystyle\tt{=\int\dfrac{1+\dfrac{1}{{t}^{2}}}{{t}^{2}+\dfrac{1}{{t}^{2}}}\,dt+\int\dfrac{1-\dfrac{1}{{t}^{2}}}{{t}^{2}+\dfrac{1}{{t}^{2}}}\,dt}$

$\sf{We\,\,know,}\\\boxed{\bf{\bullet\,\,\,\left(a\right)^2+\left(b\right)^2=\left(a+b\right)^2-2ab}}\\\boxed{\bf{\bullet\,\,\,\left(a\right)^2+\left(b\right)^2=\left(a-b\right)^2+2ab}}$

So,

$\displaystyle\tt{=\int\dfrac{1+\dfrac{1}{{t}^{2}}}{\left(t-\dfrac{1}{t}\right)^2+2\cdot t\cdot\dfrac{1}{t}}\,dt+\int\dfrac{1-\dfrac{1}{{t}^{2}}}{\left(t+\dfrac{1}{t}\right)^2-2\cdot t\cdot\dfrac{1}{t}}\,dt}$

$\displaystyle\tt{=\int\dfrac{1+\dfrac{1}{{t}^{2}}}{\left(t-\dfrac{1}{t}\right)^2+2}\,dt+\int\dfrac{1-\dfrac{1}{{t}^{2}}}{\left(t+\dfrac{1}{t}\right)^2-2}\,dt}$

$\bf{\mapsto\,\,\,\,Put\,\,\,\,t-\dfrac{1}{t}=u\,\,\,\,\,\&\,\,\,\,\,t+\dfrac{1}{t}=v}$

$\bf{\mapsto\,\,\,\,\left(1+\dfrac{1}{{t}^{2}}\right)dt=du\,\,\,\,\,\&\,\,\,\,\,\left(1-\dfrac{1}{{t}^{2}}\right)dt=dv}$

So,

$\displaystyle\tt{=\int\dfrac{du}{{u}^{2}+2}+\int\dfrac{dv}{{v}^{2}-2}}$

$\displaystyle\tt{=\int\dfrac{du}{\left(\sqrt{2}\right)^{2}+\left(u\right)^{2}}+\int\dfrac{dv}{\left(v\right)^{2}-\left(\sqrt{2}\right)^{2}}}$

$\displaystyle\tt{=tan^{-1}\left(\dfrac{u}{\sqrt{2}}\right)+\dfrac{1}{2\sqrt{2}}\,\ln\left(\dfrac{v-\sqrt{2}}{v+\sqrt{2}}\right)+C}$

Put the value of u and v,

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

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

Put the value of t,

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

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