Question

If $f\left(x\right)=\frac{x+2}{2x+3}$, then $\int {\left(\frac{f\left(x\right)}{x^2}\right)}^{1/2}dx=\frac{1}{\sqrt{2}}g\left(\frac{1+\sqrt{2f\left(x\right)}}{1-\sqrt{2f\left(x\right)}}\right)-\sqrt{\frac{2}{3}}h\left(\frac{\sqrt{3f\left(x\right)}+\sqrt{2}}{\sqrt{3f\left(x\right)}-\sqrt{2}}\right)+c$, where

• A. $g\left(x\right)={\tan }^{-1}x,h\left(x\right)=\log |x|$
• B. $g\left(x\right)=\log |x|,h\left(x\right)={\tan }^{-1}x$
• C. $g\left(x\right)=h\left(x\right)={\tan }^{-1}x$
• D. $g\left(x\right)=\log |x|,h\left(x\right)=\log |x|$

Answer 1

Answer: (A)

$g\left(x\right)={\tan }^{-1}x,h\left(x\right)=\log |x|$

$I=\int \frac{\sqrt{x+2}}{x\sqrt{2x+3}}dx$

Substitute $u=\sqrt{2x+3}\Rightarrow dx=\sqrt{2x+3}du$, use $\frac{1}{x}=\frac{2}{u^2-3},x=\frac{u^2-3}{2}:$

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

Perform $u=tan\left(v\right)\Rightarrow v=ta{n}^{-1}u,du=se{c}^{2}vdv:$

$\int \frac{se{c}^2\left(v\right)\sqrt{ta{n}^2\left(v\right)+1}}{ta{n}^2\left(v\right)-3}dv$

Simplify using $ta{n}^2\left(v\right)+1=se{c}^2\left(v\right):$

$=\int \frac{se{c}^3\left(v\right)}{ta{n}^2\left(v\right)-3}dv$

$=\int cos\left(v\right)(-\frac{1}{(si{n}^2\left(v\right)-1)(4si{n}^2\left(v\right)-3)})dv$

Substitute $w=sin\left(v\right)\Rightarrow dv=\frac{1}{cos\left(v\right)}dw:$

$=-\int \frac{1}{(w^2-1)(4w^2-3)}dw$

Factor the denominator and perform partial fraction decomposition:

$=-\int (-\frac{4}{4w^2-3}-\frac{1}{2(w+1)}+\frac{1}{2(w-1)})dw$

$=-(-4\int \frac{1}{4w^2-3}dw-\frac{1}{2}\int \frac{1}{w+1}dw+\frac{1}{2}\int \frac{1}{w-1}dw)$ ........$\left(i\right)$

Now, $\int \frac{1}{4w^2-3}dw=\int \frac{1}{(2w-\sqrt{3})(2w+\sqrt{3})}dw$

Perform partial fraction decomposition:

$=\int (\frac{1}{2.\sqrt{3}(2w-\sqrt{3})}-\frac{1}{2.\sqrt{3}(2w+\sqrt{3})})dw$

$=\frac{1}{2.\sqrt{3}}\int \frac{1}{2w-\sqrt{3}}dw-\frac{1}{2.\sqrt{3}}\int \frac{1}{2w+\sqrt{3}}dw$

$=\frac{\ln (2w-\sqrt{3})}{4.\sqrt{3}}-\frac{\ln (2w+\sqrt{3})}{4.\sqrt{3}}$

So, $-4\int \frac{1}{4w^2-3}dw-\frac{1}{2}\int \frac{1}{w+1}dw+\frac{1}{2}\int \frac{1}{w-1}dw\to$

$=-\frac{\ln (2w-\sqrt{3})}{\sqrt{3}}+\frac{\ln (2w+\sqrt{3})}{\sqrt{3}}-\frac{\ln (w+1)}{2}+\frac{\ln (w-1)}{2}$

Plug in solved integrals:

$-\int \frac{1}{(w^2-1)(4w^2-3)}dw$

$=\frac{\ln (2w-\sqrt{3})}{\sqrt{3}}-\frac{\ln (2w+\sqrt{3})}{\sqrt{3}}+\frac{\ln (w+1)}{2}-\frac{\ln (w-1)}{2}$

Undo substitution $w=sin\left(v\right):$

$=\frac{\ln (2sin\left(v\right)-\sqrt{3})}{\sqrt{3}}-\frac{\ln (2sin\left(v\right)+\sqrt{3})}{\sqrt{3}}+\frac{\ln (sin\left(v\right)+1)}{2}-\frac{\ln (sin\left(v\right)-1)}{2}$

Undo substitution $v=ta{n}^{-1}\left(u\right)$, use $sin\left(ta{n}^{-1}\left(u\right)\right)=\frac{u}{\sqrt{u^2+1}}:$

$=\frac{\ln (\frac{2u}{\sqrt{u^2+1}}-\sqrt{3})}{\sqrt{3}}-\frac{\ln (\frac{2u}{\sqrt{u^2+1}}+\sqrt{3})}{\sqrt{3}}+\frac{\ln (\frac{u}{\sqrt{u^2+1}}+1)}{2}-\frac{\ln (\frac{u}{\sqrt{u^2+1}}-1)}{2}$

Plug in solved integrals:

$\sqrt{2}\int \frac{\sqrt{u^2+1}}{u^2-3}du$

$=\sqrt{\frac{2}{3}}\ln (\frac{2u}{\sqrt{u^2+1}}-\sqrt{3})-\sqrt{\frac{2}{3}}\ln (\frac{2u}{\sqrt{u^2+1}}+\sqrt{3})+\frac{\ln (\frac{u}{\sqrt{u^2+1}}+1)}{\sqrt{2}}-\frac{\ln (\frac{u}{\sqrt{u^2+1}}-1)}{\sqrt{2}}$

Undo substitution $u=\sqrt{2x+3}\Rightarrow \frac{u}{u^2+1}=\sqrt{\frac{(2x+3)}{(2x+4)}}=\frac{1}{\sqrt{2f\left(x\right)}}$

$=\sqrt{\frac{2}{3}}\ln (\sqrt{\frac{2}{f\left(x\right)}}-\sqrt{3})-\sqrt{\frac{2}{3}}\ln (\sqrt{\frac{2}{f\left(x\right)}}+\sqrt{3})+\frac{\ln (\frac{1}{\sqrt{2f\left(x\right)}}+1)}{\sqrt{2}}-\frac{\ln (\frac{1}{\sqrt{2f\left(x\right)}}-1)}{\sqrt{2}}$

Now, taking $LCM$ inside $log$ function and then using $\ln A-\ln B=\ln \left(\frac{A}{B}\right)$:

we get, $I=\frac{1}{\sqrt{2}}\ln \left(\frac{1+\sqrt{2f\left(x\right)}}{1-\sqrt{2f\left(x\right)}}\right)-\sqrt{\frac{2}{3}}\ln \left(\frac{\sqrt{3f\left(x\right)}+\sqrt{2}}{\sqrt{3f\left(x\right)}-\sqrt{2}}\right)+c$

So, $g\left(x\right)=\ln |x|,h\left(x\right)=\ln |x|$

Hence, option $\left(D\right)$ is correct.