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Question

# ∫x2+1x4+1dx will be equal to which of the following

A

12tan1(x1x2)+c

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B

122tan1(x1x22)+c

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C

152tan1(x1x32)+c

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D

152tan1(x1x52)+c

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Solution

## The correct option is A 1√2tan−1(x−1x√2)+c To solve these kind of integration problems we have to divide the numerator as well as the denominator by x2 so that we can appropriately get the differentiation in numerator of what we substitute. Then we’ll make the suitable substitution to have a quadratic equation in the denominator. After that we can use the suitable standard formula. So, let’s divide the numerator and denominator by x2 ∫1x2+1x2+1x2dx Or ∫1x2+1(x−1x)2+2dx Now substitute t=x−1x So, dt=(1+1x2)dx ∫1x2+1(x−1x)2+2dx =∫1(t)2+2dt=∫1(t)2+(√2)2dt We can see that this is in standard form ∫1x2+a2dx which is equal to 1atan−1(xa)So,∫1(t)2+(√2)2dt=1√2tan−1(1√2) Let’s substitute t=x−1x in the above equation. 1√2tan−1(t√2)Or 1√2tan−1(x−1x√2)+c

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