Question

Find :$\int x\cdot \frac{\ell n(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$ equals :-

• A. $x\ell n(x+\sqrt{1+x^2})-x+c$
• B. $\frac{x}{2}\cdot \ell n^2(x+\sqrt{1+x^2})-\frac{x}{\sqrt{1+x^2}}+c$
• C. $\frac{x}{2}\cdot \ell n^2(x+\sqrt{1+x^2})+\frac{x}{\sqrt{1+x^2}}+c$
• D. $\sqrt{1+x^2}\ell n(x+\sqrt{1+x^2})+x+c$

Answer 1

The correct option is A $x\ell n(x+\sqrt{1+x^2})-x+c$

$\int x.\frac{ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$

Put $\sqrt{1+x^2}=t$

$\Rightarrow 1+x^2=t^2\Rightarrow 2xdx=2tdt$

$\Rightarrow xdx=tdt$

$\therefore$ $I=\int \frac{ln(\sqrt{t^2-1}+t)}{t}.tdt$

$=\int ln(\sqrt{t^2-1}+t)dt$

$=\frac{(\sqrt{t^2-1}+t)ln(\sqrt{t^2-1}+t)-(\sqrt{t^2-1}+t)+c}{(\frac{2t}{\sqrt[2]{2t^2-1}}+1)}$

$=\frac{\left(\sqrt{t^2-1}\right)[(\sqrt{t^2-1}+t)ln(\sqrt{t^2-1}+t)-(\sqrt{t^2-1}+t)]}{t+\sqrt{t^2-1}}+c$

$=\frac{x[(x+\sqrt{1+x^2})ln(x+\sqrt{1+x^2})-(x+\sqrt{1+x^2})]}{x+\sqrt{1+x^2}}+c$

$=xln(x+\sqrt{1+x^2})-x+c$ [A]