How to solve $\int \frac{4x}{\sqrt{(x-1})^3}$ by substitution?

Question

How can I solve the following integral? $$\int \frac{4x}{\sqrt{(x-1})^3}$$ As usual I tried u-substitution with $u=x-1$: $$ \int \frac{4u-4}{\sqrt{u^3}} = \int \frac{4u-4}{u^{3/2}} = \int 4u\cdot u^{-3/2} - \int4u^{-3/2} \\ = 4 \int u^{-1/2} - 4\int u^{-3/2} = 8u^{1/2} + 8u^{-1/2} + C \\ = 8\sqrt{(x-1)} + \frac{8}{\sqrt{x-1}} + C$$

The problem is, according to my textbook this answer is wrong, so what did I miss?

Answer 1

With your sub, fix your first $u$-integral by $$ \int \frac{4u+4}{u^{3/2}} \ du. $$ Now split and integrate with power rule to get $$ 8\sqrt{u} - \frac{8}{\sqrt{u}} + C. $$ Now put back in $u=x-1$ and modulo your sign mistake, I agree with you.

Answer 2

for your first integral, write $$\pm\int x\sqrt{1+x^2}dx$$ for your second integral the result is given by $$\frac{8 \left(x^2-3 x+2\right)}{\sqrt{(x-1)^3}}+C$$

Answer 3

In the enumerator, when you substitute $x$ by $u + 1$, it should yield $4u + 4$ and not $4u - 4$.

Then:

$4 ( \int u^{-1/2} du + \int u^{-3/2} du ) = 8(\sqrt u - {1 \over \sqrt u} ) + C = {8 (u - 1) \over \sqrt u} +C= {8 (x - 2) \over \sqrt {x-1}}+C$