Indefinite IntegrationeasyFree

Primitive of x·2^(ln(x²+1)) | JEE

JEE Maths question with a full step-by-step solution.

Question

A primitive of

f(x)=x\cdot 2^{\ln(x^{2}+1)}

with respect to

x

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

\dfrac{(x^{2}+1)\,2^{\ln(x^{2}+1)}}{\ln 2+1}+C

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

correct

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

Solution

Step 1: Substitute

t=x^{2}+1

, so

dt=2x\,dx\Rightarrow x\,dx=\dfrac{dt}{2}

. Step 2: Convert the exponential base. Using

2^{u}=e^{u\ln 2}

2^{\ln t}=e^{\ln 2\cdot\ln t}=\big(e^{\ln t}\big)^{\ln 2}=t^{\ln 2}

Step 3: Rewrite the integral:

I=\int x\cdot 2^{\ln(x^{2}+1)}\,dx=\int 2^{\ln t}\cdot\dfrac{dt}{2}=\dfrac{1}{2}\int t^{\ln 2}\,dt

Step 4: Integrate the power (

\ln 2

is a constant exponent):

=\dfrac{1}{2}\cdot\dfrac{t^{\ln 2+1}}{\ln 2+1}=\dfrac{t^{\ln 2+1}}{2(\ln 2+1)}

Step 5: Back-substitute

t=x^{2}+1

I=\dfrac{(x^{2}+1)^{\ln 2+1}}{2(\ln 2+1)}+C

Correct answer: (3)

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