TG EAPCET 2025 (Online) 4th May Evening Shift | Indefinite Integration Question 3 | Mathematics

$$ \begin{aligned} I_1 & =\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x, I_2 \\ & =\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x, \text { then } I_2-I_1= \end{aligned} $$

$\frac{1}{2} \log \left(\frac{e^{2 x}-e^{-2 x}+1}{e^{2 x}+e^{-2 x}-1}\right)+C$

$\frac{1}{2} \log \left(\frac{e^{2 x}-e^{-2 x}-1}{e^{2 x}+e^{-2 x}+1}\right)+C$

$\frac{1}{2} \log \left(\frac{e^{2 x}+e^{-x}+1}{e^{2 x}+e^{-x}-1}\right)+C$

$\frac{1}{2} \log \left(\frac{e^x+e^{-x}-1}{e^x+e^{-x}+1}\right)+C$

TG EAPCET 2025 (Online) 4th May Evening Shift

MCQ (Single Correct Answer)

-0

If $\int \frac{1}{x} \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=2 f(x)-2 \sin ^{-1} \sqrt{x}+c$, then $f(x)=$

$\operatorname{sech}^{-1} \sqrt{x}$

$\operatorname{cosec}^{-1} \sqrt{x}$

$\log \left(\frac{1+x}{\sqrt{x}}\right)$

$\log \left(\frac{\sqrt{1+x}-1}{\sqrt{x}}\right)$

TG EAPCET 2025 (Online) 4th May Evening Shift

MCQ (Single Correct Answer)

-0

$$ \begin{aligned} & \int \frac{3 x+2}{4 x^2+4 x+5} d x=A \log \\ & \left(4 x^2+4 x+5\right)+B \tan ^{-1}\left(\frac{2 x+1}{2}\right)+C, \text { then } A+B= \end{aligned} $$

$1 / 2$

$3 / 4$

$3 / 8$

$1 / 8$

TG EAPCET 2025 (Online) 4th May Morning Shift

MCQ (Single Correct Answer)

-0

If $\int x^3 \sin 3 x d x=\frac{1}{27}[f(x) \cos 3 x+g(x) \sin 3 x]+C$, then $f(\mathrm{l})+g(\mathrm{l})=$

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