Originally Posted by

**Glitch** **The question:**

Find $\displaystyle \int_{0}^{1} 2x(1 + x^2)^3 dx$

**My attempt:**

Let u = $\displaystyle 1 + x^2$

$\displaystyle \frac{du}{dx} = 2x$

$\displaystyle du = 2x dx$

$\displaystyle \int_{0}^{1} u^3 du = _{0}^{1}[3u^2]$** <=== This is the derivative not the integral which should be $\displaystyle \frac14 u^4$**

Find bounds in terms of u...

Sub 1 into u:

$\displaystyle 1 + (1)^2 = 2$

Sub 0 into u:

$\displaystyle 1 + 0 = 1$

$\displaystyle _{1}^{2}[3u^2]$

$\displaystyle [3(2)^2] - [3(1)^2] = 9$

However, the answer is 15/4. :/

What am I doing wrong? Thanks!