1
KCET 2023
+1
-0

$$\int \frac{1}{1+3 \sin ^2 x+8 \cos ^2 x} d x$$ is equals to

A
$$\tan ^{-1}\left(\frac{2 \tan x}{3}\right)+C$$
B
$$\frac{1}{6} \tan ^{-1}\left(\frac{2 \tan x}{3}\right)+C$$
C
$$6 \tan ^{-1}\left(\frac{2 \tan x}{3}\right)+C$$
D
$$\frac{1}{6} \tan ^{-1}(2 \tan x)+C$$
2
KCET 2022
+1
-0

$$\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x$$ is equal to

A
$$2(\sin x-x \cos \alpha)+c$$
B
$$2(\sin x+x \cos \alpha)+c$$
C
$$2(\sin x-2 x \cos \alpha)+c$$
D
$$2(\sin x+2 x \cos \alpha)+c$$
3
KCET 2022
+1
-0

If $$\int \frac{d x}{(x+2)\left(x^2+1\right)}=a \log \left|1+x^2\right|+b \tan ^{-1} x +\frac{1}{5} \log |x+2|+c,$$ then

A
$$a=\frac{-1}{10}, b=\frac{2}{5}$$
B
$$a=\frac{1}{10}, b=\frac{2}{5}$$
C
$$a=\frac{-1}{10}, b=\frac{-2}{5}$$
D
$$a=\frac{1}{10}, b=\frac{-2}{5}$$
4
KCET 2021
+1
-0

$$\int \frac{x^3 \sin \left(\tan ^{-1}\left(x^4\right)\right)}{1+x^8} d x$$ is equal to

A
$$\frac{-\cos \left(\tan ^{-1}\left(x^4\right)\right)}{4}+C$$
B
$$\frac{\cos \left(\tan ^{-1}\left(x^4\right)\right)}{4}+C$$
C
$$\frac{-\cos \left(\tan ^{-1}\left(x^3\right)\right)}{3}+C$$
D
$$\frac{\sin \left(\tan ^{-1}\left(x^4\right)\right)}{4}+C$$
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