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

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

Question

(D) Let $t = x^{10} + 10^{x}$.Differentiating both sides with respect to $x$,we get:$\frac{dt}{dx} = \frac{d}{dx}(x^{10}) + \frac{d}{dx}(10^{x})$$\fra

Vedclass · Accepted Answer

$\log \left(10^{x}+x^{10}\right)+C$

Vedclass · Answer

(D) Let $t = x^{10} + 10^{x}$.Differentiating both sides with respect to $x$,we get:$\frac{dt}{dx} = \frac{d}{dx}(x^{10}) + \frac{d}{dx}(10^{x})$$\frac{dt}{dx} = 10x^{9} + 10^{x} \log_{e} 10$Therefore,$dt = (10x^{9} + 10^{x} \log_{e} 10) dx$.Substituting these into the integral,we have:$\int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x = \int \frac{1}{t} dt$$= \log |t| + C$$= \log |x^{10} + 10^{x}| + C$Since $x^{10} + 10^{x} > 0$ for all real $x$,we can write this as $\log (x^{10} + 10^{x}) + C$.Hence,the correct option is $D$.

$\int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x$ equals

Similar Questions

$\int \frac{\tan ^4 \sqrt{x} \cdot \sec ^2 \sqrt{x}}{\sqrt{x}} d x=$

$\int \frac{3^x}{\sqrt{1-9^x}} d x=$

$\int \frac{\sin (\tan ^{-1} x)}{1+x^2} d x=$ . . . . . . $+C$.

If $\int {\frac{{\left( {2x + 3} \right)dx}}{{x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) + 1}}} = C - \frac{1}{{f(x)}}$ where $f(x)$ is of the form of $ax^2 + bx + c$ then $(a + b + c)$ equals

$ \int \frac{1}{\sqrt{x}+x \sqrt{x}} d x $

Vedclass Test Series

Exam Paper Generator

Online Exam Module