Ignore this post, found the mistake

$$\int_{1}^{4}\frac{1}{x^{(3/2)}+x}dx = \int_{1}^{4}\frac{1}{\sqrt{x}*x+x}dx$$

Subsitution:
$$t=\sqrt{x}$$
$$dt=\frac{1}{2t}dx$$

$$\int_{1}^{2}\frac{2t}{t*t^2+t}dt = 2\int_{1}^{2}\frac{1}{t^2+1}dt = 2arctan(2) - 2arctan(1)$$

Why is the above wrong?

$x=t^2$ NOT $x=t$ so you end up with

$\displaystyle \int_1^2 ~\dfrac{2t dt}{t^3 + t^2} = \int_1^2 ~\dfrac{2 dt}{t^2 + t}$

it it however fairly trivial to continue using partial fractions

Originally Posted by TriForce
Ignore this post, found the mistake

$$\int_{1}^{4}\frac{1}{x^(3/2)+x}dx = \int_{1}^{4}\frac{1}{\sqrt{x}*x+x}dx$$

Subsitution:
$$t=\sqrt{x}$$
$$dt=\frac{1}{2t}dx$$

$$\int_{1}^{2}\frac{2t}{t*t^2+t}dt = 2\int_{1}^{2}\frac{1}{t^2+1}dt = 2arctan(2) - 2arctan(1)$$

Why is the above wrong?
You replaced the x with t, but x=t^2