\int^{1}_{0}1/{(x^2+16)(x^2+25)} dx=

Correct Answer is : 19[14tan−1(14)−15tan−1(15)]

1/5[1/4tan^{-1}(1/4)-1/5tan^{-1}(1/5)]

1/9[1/4tan^{-1}(1/4)-1/5tan^{-1}(1/5)]

1/4[1/4tan^{-1}(1/4)-1/5tan^{-1}(1/5)]

1/9[1/5tan^{-1}(1/4)-1/4tan^{-1}(1/5)]

1/9[3/4tan^{-1}(1/4)-4/5tan^{-1}(1/5)]

KEAM 2015 Math Paper

Question : 13 of 120

Marks: +1, -0

∫

(x2+16)(x2+25)

dx=

1
5
[
1
4
tan−1(
1
4
)−
1
5
tan−1(
1
5
)]
1
9
[
1
4
tan−1(
1
4
)−
1
5
tan−1(
1
5
)]
1
4
[
1
4
tan−1(
1
4
)−
1
5
tan−1(
1
5
)]
1
9
[
1
5
tan−1(
1
4
)−
1
4
tan−1(
1
5
)]
1
9
[
3
4
tan−1(
1
4
)−
4
5
tan−1(
1
5
)]

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