Question

please integrate this..

Answer 1

let √(2x +3) = z
take square both sides
2x + 3 = z² .dz

differentiate
2dx = 2zdz

dx = zdz

now,
√x/√(2x +3)dx = √x/z (zdz)
=√x .dz

according to above ,
√(2x +3) = z
2x +3 = z²

2x = z² -3

x =(z²-3)/2

√x = 1/√2{√(z²-3) put this

then ,

1/√2{√(z²-3)dz }

=1/√2{ √(z²-√3²)dz}

use basic formula,
=1/√2{ z/2√(z²-3) -3/2ln(z+√z²-3)) + C

put z = √(2x +3)

=(1/2√2)(√(2x +3)) -3/2√2ln{√(2x +3)+√(2x)}

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{\int \dfrac{\sqrt{x}}{\sqrt{2x+3}}dx}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\int \dfrac{\sqrt{x}}{\sqrt{2x+3}}dx=-\dfrac{3}{2\sqrt{2}}\ln \left|\dfrac{1}{\sqrt{3}}\sqrt{2x+3}+\sqrt{\dfrac{2x}{3}}\right|+\dfrac{1}{2}\sqrt{x}\sqrt{2x+3}+C}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\text { Apply u - substitution: } u=\sqrt{2 x+3}$

$=\int \sqrt{\dfrac{u^2-3}{2}}du$

$\sqrt{\dfrac{u^2-3}{2}}=\sqrt{\dfrac{1}{2}}\sqrt{u^2-3}$

$=\int \sqrt{\dfrac{1}{2}}\sqrt{u^2-3}du$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=\sqrt{\dfrac{1}{2}}\cdot \int \sqrt{u^2-3}du$

______________________________

$\text { Apply Trig Substitution: } u=\sqrt{3} \sec (v)$

$=\sqrt{\dfrac{1}{2}}\cdot \int \:3\tan ^2\left(v\right)\sec \left(v\right)dv$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=\sqrt{\dfrac{1}{2}}\cdot \:3\cdot \int \tan ^2\left(v\right)\sec \left(v\right)dv$

$\mathrm{Use\:the\:following\:identity}:\quad \sec ^2\left(x\right)-\tan ^2\left(x\right)=1$

$\mathrm{\therefore\:}\tan ^2\left(x\right)=-1+\sec ^2\left(x\right)$

$=\sqrt{\dfrac{1}{2}}\cdot \:3\cdot \int \left(-1+\sec ^2\left(v\right)\right)\sec \left(v\right)dv$

______________________________

$\text { Expand }\left(-1+\sec ^{2}(v)\right) \sec (v):-\sec (v)+\sec ^{3}(v)$

$=\sqrt{\dfrac{1}{2}}\cdot \:3\cdot \int \:-\sec \left(v\right)+\sec ^3\left(v\right)dv$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=\sqrt{\dfrac{1}{2}}\cdot \:3\left(-\int \sec \left(v\right)dv+\int \sec ^3\left(v\right)dv\right)$

______________________________

$\begin{array}{l}\int \sec (v) d v=\ln |\tan (v)+\sec (v)| \\\\\int \sec ^{3}(v) d v=\dfrac{1}{2} \sec (v) \tan (v)+\dfrac{1}{2} \ln |\tan (v)+\sec (v)|\end{array}$

$\sf{=\sqrt{\dfrac{1}{2}}\cdot \:3\left(-\ln \left|\tan \left(v\right)+\sec \left(v\right)\right|+\dfrac{1}{2}\sec \left(v\right)\tan \left(v\right)+\dfrac{1}{2}\ln \left|\tan \left(v\right)+\sec \left(v\right)\right|\right)}$

______________________________

$\text { Substitute back }$

$=\sqrt{\dfrac{1}{2}}\cdot \:3\left(-\ln \left|\tan \left(\arcsec \left(\dfrac{1}{\sqrt{3}}\sqrt{2x+3}\right)\right)+\sec \left(\arcsec \left(\dfrac{1}{\sqrt{3}}\sqrt{2x+3}\right)\right)\right|+$

$\dfrac{1}{2}\sec \left(\arcsec \left(\dfrac{1}{\sqrt{3}}\sqrt{2x+3}\right)\right)\tan \left(\arcsec \left(\dfrac{1}{\sqrt{3}}\sqrt{2x+3}\right)\right)$$+\dfrac{1}{2}\ln \left|\tan \left(\arcsec \left(\dfrac{1}{\sqrt{3}}\sqrt{2x+3}\right)\right)+\sec \left(\arcsec \left(\dfrac{1}{\sqrt{3}}\sqrt{2x+3}\right)\right)\right|\right)$

$\sf{Simplify\:\:\sqrt{\dfrac{1}{2}}\cdot \:3\left(-\ln \left|\tan \left(\arcsec \left(\dfrac{1}{\sqrt{3}}\sqrt{2x+3}\right)\right)+\sec \left(\arcsec \left(\dfrac{1}{\sqrt{3}}\sqrt{2x+3}\right)\right)\right|}$

$\sf{+\dfrac{1}{2}\sec \left(\arcsec \left(\dfrac{1}{\sqrt{3}}\sqrt{2x+3}\right)\right)\tan \left(\arcsec \left(\dfrac{1}{\sqrt{3}}\sqrt{2x+3}\right)\right)}$$\sf{+\dfrac{1}{2}\ln \left|\tan \left(\arcsec \left(\dfrac{1}{\sqrt{3}}\sqrt{2x+3}\right)\right)+\sec \left(\arcsec \left(\dfrac{1}{\sqrt{3}}\sqrt{2x+3}\right)\right)\right|\right)}$

$\sf{=-\dfrac{3}{2\sqrt{2}}\ln \left|\dfrac{1}{\sqrt{3}}\sqrt{2x+3}+\sqrt{\dfrac{2x}{3}}\right|+\dfrac{1}{2}\sqrt{x}\sqrt{2x+3}}}$

$\bf{Add\:a\:constant\:to\:the\:solution}$

$\boxed{\sf{=-\frac{3}{2\sqrt{2}}\ln \left|\frac{1}{\sqrt{3}}\sqrt{2x+3}+\sqrt{\frac{2x}{3}}\right|+\frac{1}{2}\sqrt{x}\sqrt{2x+3}+C}}$