The primitive of f(x) = x cdot 2^{ln(x^2 + 1) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I = \int x \cdot 2^{\ln(x^2 + 1)} dx$.Substitute $t = x^2 + 1$,then $dt = 2x dx$,which implies $x dx = \frac{dt}{2}$.The integral becomes $I

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int x \cdot 2^{\ln(x^2 + 1)} dx$.Substitute $t = x^2 + 1$,then $dt = 2x dx$,which implies $x dx = \frac{dt}{2}$.The integral becomes $I = \frac{1}{2} \int 2^{\ln t} dt$.Using the property $a^{\ln b} = b^{\ln a}$,we have $2^{\ln t} = t^{\ln 2}$.So,$I = \frac{1}{2} \int t^{\ln 2} dt$.Integrating with respect to $t$,we get $I = \frac{1}{2} \cdot \frac{t^{\ln 2 + 1}}{\ln 2 + 1} + C$.Substituting back $t = x^2 + 1$,we get $I = \frac{(x^2 + 1)^{\ln 2 + 1}}{2(\ln 2 + 1)} + C$.

The primitive of $f(x) = x \cdot 2^{\ln(x^2 + 1)}$ with respect to $x$ is:

Similar Questions

$\int (x + 3)({x^2} + 6x + 10)^9 \, dx$ equals

Let $f(x) = \frac{x}{(1+x^n)^{1/n}}$ for $n \geq 2$ and $g(x) = \underbrace{(f \circ f \circ \ldots \circ f)}_{n \text{ times }}(x)$. Then $\int x^{n-2} g(x) \, dx$ equals

$\int \frac{1}{(\cos^{-1} x) \sqrt{1 - x^2}} dx = $

$\int \frac{3x^2}{\sqrt{9 - 16x^6}} \, dx = $

$\int \left[ \frac{1+\log x}{\cos^{2}(x \log x)} \right] dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module