Question

int(1)/((x+1)sqrt(2x^(2)+3x+1))dx​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\displaystyle\int\dfrac{1}{(x+1)\sqrt{2x^2+3x+1}}\,dx}$

$\sf{Let\,\,\,x+1=\dfrac{1}{t}}$

$\sf{\implies\,dx=-\dfrac{1}{t^2}\,dt}$

So,

$\sf{\displaystyle\int\dfrac{1}{\dfrac{1}{t}\,\sqrt{2\left(\dfrac{1}{t}-1\right)^2+3\left(\dfrac{1}{t}-1\right)+1}}\,\cdo\dfrac{-1}{t^2}\,dt}$

$\sf{=-\displaystyle\int\dfrac{t}{t^2\,\sqrt{2\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+3\left(\dfrac{1}{t}-1\right)+1}}\,dt}$

$\sf{=-\displaystyle\int\dfrac{1}{t\,\sqrt{\dfrac{2}{t^2}-\dfrac{4}{t}+2+\dfrac{3}{t}-3+1}}\,dt}$

$\sf{=-\displaystyle\int\dfrac{1}{t\,\sqrt{\dfrac{2}{t^2}-\dfrac{1}{t}}}\,dt}$

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

$\sf{=-\displaystyle\int\dfrac{1}{t\cdot\dfrac{1}{t}\sqrt{2-t}}\,dt}$

$\sf{=-\displaystyle\int\dfrac{1}{\sqrt{2-t}}\,dt}$

$\sf{=2\sqrt{2-t}+C}$

$\sf{=2\sqrt{2-\dfrac{1}{x+1}}+C}$

$\sf{=2\sqrt{\dfrac{2(x+1)-1}{x+1}}+C}$

$\sf{=2\sqrt{\dfrac{2x+1}{x+1}}+C}$

Answer 2

Given that :-

$\int \frac{1}{(x+1) \sqrt{2 x^{2}+3 x+1}} d x$

As we have to solve the given integration,

Let,

$x+1=\frac{1}{t}$

$\Longrightarrow \mathrm{d} \mathrm{x}=-\frac{1}{\mathrm{t}^{2}} \mathrm{dt}$

So, it is written as,

$= \int \frac{1}{\frac{1}{t} \sqrt{2\left(\frac{1}{t}-1\right)^{2}+3\left(\frac{1}{t}-1\right)+1}} \frac{-1}{t^{2}} d t$

$=-\int \frac{\mathrm{t}}{\mathrm{t}^{2} \sqrt{2\left(\frac{1}{\mathrm{t}^{2}}-\frac{2}{\mathrm{t}}+1\right)+3\left(\frac{1}{\mathrm{t}}-1\right)+1}} \mathrm{dt}$

$=-\int \frac{1}{t \sqrt{\frac{2}{\mathrm{t}^{2}}-\frac{4}{\mathrm{t}}+2+\frac{3}{\mathrm{t}}-3+1}} \mathrm{dt}$

By further simplifying it we get,

$=-\int \frac{1}{\mathrm{t} \sqrt{\frac{2}{\mathrm{t}^{2}}-\frac{1}{\mathrm{t}}}} \mathrm{dt}$

$=-\int \frac{1}{t \sqrt{\frac{1}{t^{2}}(2-t)}} d t$

$=-\int \frac{1}{\sqrt{2-\mathrm{t}}} \mathrm{dt}$

Putting t as $\frac{1}{x+1}$ we get,

$=2 \sqrt{2-\mathrm{t}}+\mathrm{C}$

$=2 \sqrt{2-\frac{1}{x+1}}+C$

$=2 \sqrt{\frac{2(x+1)-1}{x+1}}+C$

$=2 \sqrt{\frac{2 x+1}{x+1}}+C$