Primitive of f (x) = x , , 2^ ,( x^2 + 1) w.r.t. x is - Vidyadip

MCQ

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

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

✓

Answer

Correct option: C.

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

c
$I = \int {x\,{2^{\ln \,({x^2} + 1)}}dx} $

let $x^2 + 1 = t$ ; $x\, dx = \frac{{dt}}{2}$
Hence $I = \frac{1}{2}\int {{2^{\ln \,t}}dt} $

$=\frac{1}{2}\int {{t^{\ln \,2}}dt} $

$= \frac{1}{2}\,\cdot\,\frac{{{t^{\ln \,2 + 1}}}}{{\ln 2 + 1}} + C$

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

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