Question

# ∫x2(x2+1)(x2+4)dx is equal to

A

13tan1x+23tan1x2+C
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B

13tan1x23tan1x2+C
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C

13tan1x+23tan1x2+C
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D
13tan1x23tan1x2+C
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Solution

## The correct option is A −13tan−1x+23tan−1x2+CThese kind of questions can be solved easily using integration by partial fraction. Here, we can write our integrand as x2(x2+1)(x2+4)=ax+bx2+1+cx+dx2+4 ⇒x2(x2+1)(x2+4)= (ax+b)(x2+4)+(cx+d)(x2+1)(x2+1)(x2+4) ⇒x2=(ax+b)(x2+4) +(cx+d)(x2+1)⇒x2=(a+c)x3+(d+b)x2 +(4a+c)x+(4b+d) Now, equating the co-efficients of different powers of x we get the equations: a+c=0; 4a+c=0 and b+d=1; 4b+d=0 Now solving these two sets of equations we get: a=0, c=0 and b=−13 , d=43 Now, we can write our integral as ∫x2(x2+1)(x2+4)dx =13∫(4x2+4−1x2+1]dx we know that ∫1x2+a2dx=1atan−1(xa)+C So, we get integral as: I=−13tan−1(x)+43×12tan−1(x2) +C⇒I= −13tan−1(x)+23tan−1(x2)+C

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