<s>[INST] Where is the mistake when finding $\int x \sqrt{4+5 x} \ dx$? 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?
 [/INST] 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:
$\int{\frac{u-4}{5}(\sqrt{u})(\frac{1}{5}) du}$
</s>