int {frac{{10{x^9} + {{10}^x}{{log }_e}10}}{{ | vedclass

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

Question

(B) Let $I = \int {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{{10}^x} + {x^{10}}}}} \;dx$. Substitute $t = {x^{10}} + {10^x}$. Differentiating with re

Vedclass · Accepted Answer

$\log ({10^x} + {x^{10}}) + c$

Vedclass · Answer

(B) Let $I = \int {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{{10}^x} + {x^{10}}}}} \;dx$. Substitute $t = {x^{10}} + {10^x}$. Differentiating with respect to $x$,we get $dt = (10{x^9} + {10^x}{\log _e}10) \;dx$. Substituting these into the integral,we get $I = \int {\frac{1}{t}} \;dt$. Integrating,we get $I = \log |t| + c$. Substituting back the value of $t$,we get $I = \log |{x^{10}} + {10^x}| + c$.

$\int {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{{10}^x} + {x^{10}}}}} \;dx = $

Similar Questions

The integral $\int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{4 x^{2}-4 x+6}} d x$ is equal to (where $c$ is a constant of integration)

$\int \frac{e^{x}(1+x)}{\cos ^{2}(x e^{x})} d x$ equals

$\int \left[ \frac{x^4-x}{x^{20}} \right]^{1/4} dx =$

Find $\int \frac{(x^{4}-x)^{\frac{1}{4}}}{x^{5}} dx$

Let $I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x$. If $I(0)=3$,then $I\left(\frac{\pi}{12}\right)$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module