Help with $\int \frac{1}{\sqrt{x\cdot(1 - 3\sqrt{x})}}\ \text{d}x$ by substitution.

Question

$$\int \frac{1}{\sqrt{x} (1 - 3\sqrt{x})}$$

I tried with the substitution $u = 1-3\sqrt{x}$

I am confused with how to finish this problem I know I am supposed to substitute $u$ and $\text{d}u$ in but I am not sure how to finish it.

Answer 1

Let's set the problem, then you may proceed by yourself.

Performing the substitution

$$u = 1 - 3\sqrt{x} ~~~ \text{and so} ~~~ \sqrt{x} = -\frac{u-1}{3}$$

Hence

$$\text{d}u = -\frac{3}{2\sqrt{x}} = -\frac{3}{2}\frac{-3}{u-1}\ \text{d}x$$

$$\text{d}u = \frac{9}{2(u-1)}\ \text{d}x ~~~~~ \to ~~~~~ \text{d}x = \frac{2(u-1)}{9}\ \text{d}u$$

Then your integral becomes

$$\int -\frac{1}{\frac{u-1}{3}u}\frac{2(u-1)}{9} = -\int \frac{2}{3u}\ \text{d}u$$

Now you can proceed for sure.

Answer 2

Hint: Try $u=\sqrt x$, $du=\frac{1}{2\sqrt x}$

Answer 3

The substitution is a good one. If $u=1-3x^{1/2}$, then

$$du=\frac12(-3x^{-1/2})dx=-\frac3{2\sqrt x}dx$$

$$-\frac23\int\frac1{1-3\sqrt x}\left(-\frac3{2\sqrt x}dx\right)=-\frac23\int\frac{du}u$$