∫1x+1√x−2dx 2√3tan−1√x+1√3+C 2√3tan−1√x−2√3+C 2√3tan−1√
| ∫1(x+1)√x−2dx
A. 2√3(tan−1(√x+1√3))+C
B. 2√3(tan−1(√x−2√3))+C
C. 2√3(tan−1(√x−2√2))+C
D. 1√2(tan−1(√x−2√2))+C
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Right Answer is: B
SOLUTION
The given form is one of the forms of irrational algebraic functions which is ∫1(ax+b)√cx+ddx.
Now to evaluate such kind of integrals we need to substitute cx+d=t2
So, here we’ll substitute x−2=t2
& dx = 2t.dt
So the integral becomes
∫2t.(t2+3)t.dt
=∫2(t2+3)dt (use the standard formulae)
=2√3(tan−1(t√3))+C
Or 2√3(tan−1(√x−2√3))+C (Substituting t = √x−2)