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a
15log|x+2|−110logx2+1+25tan−1x+C
b
log|x+2|+110logx2+1+15tan−1x+C
c
5log|x+2|+10logx2+1+tan−1x+C
d
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Correct option is A
Let 1(x+2)x2+1=Ax+2+Bx+Cx2+1 ⇒ 1=Ax2+1+(Bx+C)(x+2) Put x=−2, we get A=15Now, comparing the coefficients of x2 and constant term, we get 0=A+Band 1=A+2C⇒ B=−15,C=25So, I=15∫dxx+2−15∫xx2+1dx+25∫dxx2+1=15log|x+2|−110logx2+1+25tan−1x+CSimilar Questions
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