1
TG EAPCET 2025 (Online) 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\frac{x+3}{(x+1)\left(x^2+2\right)}=\frac{a}{x+1}+\frac{b x+c}{x^2+2}$, then $a-b+c=$

A

0

B

1

C

3

D

2

2
TG EAPCET 2025 (Online) 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int e^{-x}\left(x^3-2 x^2+3 x-4\right) d x= $$

A

$-e^{-x}\left(x^3-x^2+5 x-1\right)+C$

B

$e^{-x}\left(x^3-x^2+5 x-1\right)+C$

C

$e^{-x}\left(x^3+x^2+5 x+1\right)+C$

D

$-e^{-x}\left(x^3+x^2+5 x+1\right)+C$

3
TG EAPCET 2025 (Online) 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int\left(1+\tan ^2 x\right)(1+2 x \tan x) d x= $$

A

$x \sec x+C$

B

$x \tan ^2 x+C$

C

$x \sec ^2 x+C$

D

$x \tan x+C$

4
TG EAPCET 2025 (Online) 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

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

A

$\frac{\left(\tan ^{-1} x\right)^2}{4}-\frac{x \tan ^{-1} x}{2\left(1+x^2\right)}+\frac{1-x^2}{4\left(1+x^2\right)}+C$

B

$\frac{\left(\tan ^{-1} x\right)^2}{4}-\frac{4 x \tan ^{-1} x+1-x^2}{8\left(1+x^2\right)}+C$

C

$\frac{\left(\tan ^{-1} x\right)^2}{4}-\frac{x \tan ^{-1} x}{\left(1+x^2\right)}-\frac{1-x^2}{4\left(1+x^2\right)}+C$

D

$\frac{(\tan x)^2}{4}+\frac{4 x \tan ^{-1} x-1+x^2}{4\left(1+x^2\right)}+C$

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