$\int\limits_{0}^{32/9}\sqrt{1+\frac{9x}{4}}dx$

Question

Question : Solve $\int\limits_{0}^{32/9}\sqrt{1+\frac{9x}{4}}dx$

My Try: Let u = $1+\frac{9x}{4}$

Then,

$$du = \frac{9x}{4}dx$$

$$dx = \frac{4du}{9}$$

Substituting the above in the main equation, we get:

$$\int\limits_{0}^{32/9} \sqrt u \left(\frac{4du}{9}\right)$$

$$ \left. \left(\frac{4}{9}\right)\frac{2(u^{3/2})}{3}\right|_{0}^{\frac{32}{9}}$$

$$ \frac{4}{9} \frac{2\left(\left(1+\frac{9(32/9)}{4}\right)^{3/2}\right)}{3}$$

$$ \left(\frac{4}{9}\right)\left(\frac{2}{3}\right)\left(9^{\frac{3}{2}}\right)$$

= 8

But the correct answer is 7.703...

What am I doing wrong?

EDIT : Thank you all for your help. In case anyone else has this problem this website builds upon Oliver's answer.

Answer 1

Performing the change of variable $\displaystyle u=1+\frac{9x}{4}$ gives $$ u(0)=1,\quad u(32/9)=9,\quad du=\frac94dx,\quad dx=\frac49du,\quad $$ then you just get $$ \begin{align} \int_{0}^{32/9}\sqrt{1+\frac{9x}{4}}dx&=\frac49\int_{1}^{9}\sqrt{u}\:du\\\\ &=\frac49\left[\frac{2}{3}u^{3/2} \right]_1^9\\\\ &=\frac{8}{27}\left(9^{3/2}-1^{3/2} \right)\\\\ &=\frac{208}{27}. \end{align} $$