Where is the mistake when finding $\int x \sqrt{4+5 x} \ dx$?

Question

We have to find the integral of $\frac{dy}{dx}=x \sqrt{4+5x}$. This is from Morris Kline's book, Chapter $7$, exercise $5$, question $1$m.

I try to solve like following:

Let $u=4+5x$. Then $x=\frac{u-4}{5}$. Hence $\frac{d y}{d x}=\left(\frac{u-4}{5}\right) \sqrt{u}$

Now $$\int(x \sqrt{4+5 x}) \cdot d x=\int\left(\frac{u-4}{5}\right) \sqrt{u} \cdot d x$$ $$ = \int\left(\frac{u^{3 / 2}-4 u^{1 / 2} }{5}\right) \cdot d x$$

$$=\frac{\int u^{3 / 2}-4 \int u^{1 / 2}}{5} \cdot d x$$

$$=\frac{\frac{u^{5 / 2}}{5 / 2} -\frac{4 u^{3 / 2}}{3 / 2}}{5} + C$$

$$=\frac{2 u^{5 / 2}}{25}-\frac{8 u^{3 / 2}}{15}+C$$

$$\Rightarrow y=\frac{2(4+5 x)^{5 / 2}}{25} - \frac{8(4+5 x)^{3 / 2}}{15}+C$$

Where $C$ is a constant.

But the answer given in the book is: $$y=\left(\frac{2(4+5 x)^{5 / 2}}{125}\right)-\left(\frac{8(4+5 x)^{3 / 2}}{75}\right)+C.$$

Where is the mistake?

Answer 1

Remember change of variables requires substituting the differentiable:

$x = \frac{u - 4}{5} \rightarrow dx = \frac{1}{5} du$

So really after substitution you should have:

$\int{\frac{u-4}{5}(\sqrt{u})(\frac{1}{5}) du}$

Answer 2

Answer:

$I=\int_{}^{}  x\sqrt{5x+4}$ $\Rightarrow $ $I=\frac{1} {5}\int_{}^{}  5x\sqrt{5x+4}$ $\Rightarrow $ $I=\frac{1} {5}\int_{}^{}  (5x+4-4)\sqrt{5x+4}$

$\Rightarrow $ $I=\frac{1} {5}\int_{}^{}  (5x+4)^{\frac{3}{2}} -\frac{4}{25}\int_{}^{} 5\sqrt{5x+4} $

$\Rightarrow $$I= \frac{1}{5} [\frac{2}{25} (5x+4)^{\frac{5}{2}}] - \frac{4}{25} [\frac{2} {3} (5x+4)^{\frac{3}{2}} ] +c$

I=$\frac{2}{125} [(5x+4)^{\frac{5}{2}}] - \frac{8}{75} [ (5x+4)^{\frac{3}{2}} ]+c$.