int left( frac{(x^2+2) a^{(x+tan^{-1} x)}}{x^ | vedclass

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

Question

(C) Let $I = \int \frac{(x^2+2) a^{(x+	an^{-1} x)}}{x^2+1} dx$. We can rewrite the integrand as: $I = \int \left( \frac{x^2+1+1}{x^2+1} ight) a^{(x

Vedclass · Accepted Answer

$\frac{a^{x+	an^{-1} x}}{\log a}+c$

Vedclass · Answer

(C) Let $I = \int \frac{(x^2+2) a^{(x+	an^{-1} x)}}{x^2+1} dx$. We can rewrite the integrand as: $I = \int \left( \frac{x^2+1+1}{x^2+1} ight) a^{(x+	an^{-1} x)} dx = \int \left( 1 + \frac{1}{x^2+1} ight) a^{(x+	an^{-1} x)} dx$. Let $u = x + 	an^{-1} x$. Then,$du = (1 + \frac{1}{1+x^2}) dx$. Substituting these into the integral: $I = \int a^u du$. Using the standard integral formula $\int a^u du = \frac{a^u}{\ln a} + C$,we get: $I = \frac{a^u}{\ln a} + C$. Substituting back $u = x + 	an^{-1} x$: $I = \frac{a^{x+	an^{-1} x}}{\ln a} + C$.

$\int \left( \frac{(x^2+2) a^{(x+\tan^{-1} x)}}{x^2+1} \right) dx = $ . . . . . .

Similar Questions

The integral $\int {\frac{{dx}}{{{{(x + 1)}^{\frac{3}{4}}}{{(x - 2)}^{\frac{5}{4}}}}}} $ is equal to

The integral $\int \frac{e^{3 \log_{e} 2x} + 5e^{2 \log_{e} 2x}}{e^{4 \log_{e} x} + 5e^{3 \log_{e} x} - 7e^{2 \log_{e} x}} dx$,where $x > 0$,is equal to ....... (where $c$ is a constant of integration).

If $0 < x < 1$ and $\int \frac{dx}{\sqrt{x^2-x^5}} = \frac{1}{3} \log |f(x)| + C$,then $f\left(\frac{1}{2}\right) = $

Integrate the function: $\frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}}$

$\int \frac{1}{e^x+1} \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module