Question
If ∫(x−1x+1)dx√x3+x2+x=2tan−1√f(x)+C, find f(x).
Answer: Option A
:
A
I=∫(x−1)dx(x+1)x√x+1+1x=∫(x−1)(x+1)dx(x+1)2x√x+1+1x=∫(1−1x2)dx(x+1x+2)√x+1x+1Putx+1+1x=t2(1−1x2)dx=2tdt=∫2tdt(t2+1)t=2tan−1t+c=2tan−1(√x+1x+1)+c
f(x) = x+1x+1
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:
A
I=∫(x−1)dx(x+1)x√x+1+1x=∫(x−1)(x+1)dx(x+1)2x√x+1+1x=∫(1−1x2)dx(x+1x+2)√x+1x+1Putx+1+1x=t2(1−1x2)dx=2tdt=∫2tdt(t2+1)t=2tan−1t+c=2tan−1(√x+1x+1)+c
f(x) = x+1x+1
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