How do you evaluate the integral ∫x√x−1−√x?

1 Answer
mason m
Jan 29, 2017

−25(x−1)52−23(x−1)32−25x52+C

Explanation:

Multiply first by the conjugate of the denominator to simplify the integrand.

I=∫x(√x−1+√x)(√x−1−√x)(√x−1+√x)dx

The denominator is now in the form (a−b)(a+b)=a2−b2, where a=√x−1 and b=x.

I=∫x(√x−1+√x)(x−1)−xdx

I=−∫x(√x−1+√x)dx

Distributing, splitting up the integral and rewriting:

I=−∫x(x−1)12dx−∫x32dx

The second integral can be directly integrated using ∫xndx=xn+1n+1+C, where n≠−1.

I=−∫x(x−1)12dx−x5252

I=−∫x(x−1)12dx−25x52

For the remaining integral, let u=x−1. This implies that x=u+1 and du=dx.

I=−∫(u+1)u12du−25x52

Distributing (u+1)u12 into separate integrals:

I=−∫u32du−∫u12du−25x52

Using the rule from earlier:

I=−u5252−u3232−25x52+C

From u=x−1:

I=−25(x−1)52−23(x−1)32−25x52+C

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