TS EAMCET 2020 (Online) 14th September Morning Shift | Indefinite Integration Question 109 | Mathematics

$$ \begin{aligned} & \int \frac{\left(3 x^3+\left(1-30 x^2\right) f(x)\right)}{(10 f(x)-x)\left(x^3-f(x)\right)^2} d x \\ & =\frac{A}{B x^3+D f(x)}+C \text { then } A+B+D= \end{aligned} $$

$\frac{1}{2}$

-1

TS EAMCET 2020 (Online) 14th September Morning Shift

MCQ (Single Correct Answer)

-0

If $\int x[\log (1+x)]^3 d x=\frac{(1+x)^2}{16}(f(x))+(1+x)(g(x))$, then

$$ f(x)+g(x)= $$

$\log (1+x)\left[6+9(\log (1+x))-7(\log (1+x))^2\right]+C$

$\log (1+x) x^3+7(\log (1+x))^2+4 \log (1+x)+C$

$12-18 \log (1+x)+15(\log (1+x))^2-9(\log (1+x))^3+C$

$6 \log (1+x)-9(\log (1+x))^2+7(\log (1+x))^3+C$

TS EAMCET 2020 (Online) 14th September Morning Shift

MCQ (Single Correct Answer)

-0

$$ \int \frac{\left(x+\sqrt{1+x^2}\right)^2}{\sqrt{1+x^2}} d x= $$

$\frac{x}{\sqrt{1+x^2}}+C$

$\log \left|x+\sqrt{1+x^2}\right|+C$

$x+\sqrt{1+x^2}+C$

$\frac{\left(x+\sqrt{1+x^2}\right)^2}{2}+C$

TS EAMCET 2020 (Online) 14th September Morning Shift

MCQ (Single Correct Answer)

-0

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

$\frac{4}{15}\left(\frac{\left(x^3-1\right)^5}{x^{15}}\right)^{1 / 4}+C$

$\frac{4}{15}\left(\frac{x^4+1}{x^4}\right)^{1 / 4}+C$

$\frac{\sqrt{x^4+x^2+1}}{x}+C$

$\frac{3}{4}\left(x^{4 / 3}+x^{1 / 3}\right)+C$

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