1
AP EAPCET 2025 - 23rd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $\int \frac{1}{\left((x+4)^3(x+1)^5\right)^{1 / 4}} d x=A \cdot\left(\frac{x+4}{x+1}\right)^n+C$

A

$n$ A=3

B

$n+\frac{1}{A}=-\frac{1}{2}$

C

$A+n=1$

D

$A=n$

2
AP EAPCET 2025 - 23rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

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

A

$\frac{1}{3} \log \left(\frac{x+1}{x^2+x+1}\right)+C$

B

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

C

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

D

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

3
AP EAPCET 2025 - 23rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int \frac{x^4-16 x^2+2 x+8}{x^3-4 x^2+2} d x= $$

A

$\frac{x^2+8 x+C}{2}$

B

$x^2+8 x+C$

C

$x^3-4 x+C$

D

$\frac{x^2-8 x+C}{2}$

4
AP EAPCET 2025 - 23rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int \frac{\sec ^2 x}{(\sec x+\tan x)^{\frac{5}{2}}} d x= $$

A

$-\frac{(\sec x+\tan x)^{\frac{5}{2}}}{5}-\frac{(\sec x+\tan x)^{\frac{7}{2}}}{7}+C$

B

$-\frac{(\sec x-\tan x)^{\frac{5}{2}}}{5}-\frac{(\sec x-\tan x)^{\frac{7}{2}}}{7}+C$

C

$-\frac{(\sec x+\tan x)^{\frac{3}{2}}}{3}-\frac{(\sec x+\tan x)^{\frac{7}{2}}}{7}+C$

D

$-\frac{(\sec x-\tan x)^{\frac{3}{2}}}{3}-\frac{(\sec x-\tan x)^{\frac{7}{2}}}{7}+C$

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