Primitive of fx-x -2Inx2-1 with respect to x isA- x2-1In 2-1-2In 2-1-CB- 2lnx2-1-x2-1-CC- x2-1 2lnx2-1-x2-1ln 2 1-CD- x 2-1 In 2-2In 2-1- C

The correct option is A (x^2+1)^In 2+1/2(In 2+1)+CI=∫ x2^In(x^2+1)dx let x^2+1 =t; x dx =dt/2 Hence I=1/2∫ 2^In tdt =1/2∫ t^In 2dt =1/2. t^In 2+1/In 2+1+C =1/2 ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

Primitive of ...

Question

Primitive of $f (x) = x {.2}^{I n (x^{2} + 1)}$ with respect to x is

\frac{2^{I n (x^{2} + 1)}}{(x^{2} + 1)} + C

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\frac{(x^{2} + 1) 2^{I n (x^{2} + 1)}}{I n 2 + 1} + C

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\frac{(x^{2} + 1)^{I n 2 + 1}}{2 (I n 2 + 1)} + C

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\frac{(x^{2} + 1)^{I n 2}}{2 (I n 2 + 1)} + C

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Solution

The correct option is C $\frac{(x^{2} + 1)^{I n 2 + 1}}{2 (I n 2 + 1)} + C$
$I = \int x 2^{I n (x^{2} + 1)} d x$ let $x^{2} + 1 = t; x d x = \frac{d t}{2}$
Hence $I = \frac{1}{2} \int 2^{I n t} d t = \frac{1}{2} \int t^{I n 2} d t = \frac{1}{2} . \frac{t^{I n 2 + 1}}{I n 2 + 1} + C = \frac{1}{2} . \frac{(x^{2} + 1)^{I n 2 + 1}}{I n 2 + 1} + C$

Suggest Corrections

Primitive of f(x)=x.2In(x2+1) with respect to x is

The correct option is C (x2+1)In2+12(In2+1)+CI=∫x2In(x2+1)dx let x2+1=t;xdx=dt2 Hence I=12∫2Intdt=12∫tIn2dt=12.tIn2+1In2+1+C=12.(x2+1)In2+1In2+1+C

Primitive of $f (x) = x {.2}^{I n (x^{2} + 1)}$ with respect to x is