Simplifying Elliptic Integrals

Question

According to wolfram alpha, the following equation holds.
$$\int_0^1 \frac{ku^2}{\sqrt{(1 - u^2) (1 - k u^2)}}=K(k)-E(k)$$
where $K$ and $E$ are complete elliptic integrals of the first and second kind, respectively.
It looked so easy that I tried to prove it myself, but without success.
Can someone please explain this to me?

Answer 1

$$\begin{align*} I &= \int_0^1 \frac{ku^2}{\sqrt{(1 - u^2) (1 - k u^2)}}\,du \\ &= \int_0^1 \frac{1}{\sqrt{(1 - u^2) (1 - k u^2)}}\,du-\int_0^1 \frac{1-ku^2}{\sqrt{(1 - u^2) (1 - k u^2)}}\,du \\ &= \int_0^1 \frac{du}{\sqrt{(1 - u^2) (1 - k u^2)}}\,du-\int_0^1 \frac{\sqrt{1-ku^2}}{\sqrt{1 - u^2}}\,du \\ &= K(k)-E(k) \end{align*}$$

per Wolfram|Alpha's convention for $K$ and $E$.