Question

Primitive of $f\left(x\right)=x{.2}^{\ln (x^2+1)}$ w.r.t $x$ is

• A. $\frac{2^{\ln (x^2+1)}}{2(x^2+1)}+C$
• B. $\frac{(x^2+1)2^{\ln (x^2+1)}}{\ln 2+1}+C$
• C. $\frac{(x^2+1{)}^{(\ln 2+1)}}{2(\ln 2+1)}+C$
• D. $\frac{(x^2+1{)}^{\ln 2}}{2(\ln 2+1)}+C$

Answer 1

Answer: (C)

$\frac{(x^2+1{)}^{(\ln 2+1)}}{2(\ln 2+1)}+C$

By logarithmic property,$2^{ln(x^2+1)}=(x^2+1{)}^{ln2}$
$=>\int x.(x^2+1{)}^{ln2}dx$
Let,
$x^2+1=t$

$=>2xdx=dt=>xdx=\frac{dt}{2}$
$=>\frac{1}{2}\int (t{)}^{ln2}dt$
$=\frac{t^{(ln2+1)}}{2(ln2+1)}+c$
$=\frac{(x^2+1{)}^{(ln2+1)}}{2(ln2+1)}+c$