Definite integral $\int_{0}^{1} x\sqrt{1+2x} \,\mathrm dx$

Question

I've tried to solve this integral by parts but I keep getting the wrong result. Please help me spot my mistake.

$$\int x\sqrt{1+2x}\,\mathrm dx=x\int\sqrt{1+2x}\,\mathrm dx-\int\sqrt{1+2x}\,\mathrm dx=(x-1)\int\sqrt{1+2x}\,\mathrm dx$$ Then I make substitution $u=1+2x, \mathrm du=2\mathrm dx$ : $$\int\sqrt{1+2x}\,\mathrm dx=\frac{1}{2}\int\sqrt{u}\,\mathrm du=\frac{1}{3}u^{\frac{3}{2}}.$$ Putting it all back together $$\int_{0}^{1} x\sqrt{1+2x}\,\mathrm dx=\left((x-1)\frac{1}{3}(1+2x)^{\frac{3}{2}}\right)|_0^1=\frac{1}{3}.$$

However, the correct solution should be (at least according to wolframalpha) $\dfrac{1+6\sqrt{3}}{15}$.

Answer 1

You did the IBP wrong, you are missing an integral. Indeed, with $u= x, v= \int \sqrt{1+2x} dx$ then IBP yields

$$\int x \sqrt{1+2x}dx = \int u dv = uv -\int v du = x\int\sqrt{1+2x}dx-\int\left(\int\sqrt{1+2x}dx\right) dx$$

Also, it is easier to do the substitution $u=1+2x$ from the beginning.

Answer 2

Your first line is incorrect.

Let $u=x$ and $dv=\sqrt{1+2x}.$

Then the integral can be rewritten as

$$x\dfrac{1}{3}\left(1+2x\right)^{3/2}\biggr|_0^1-\dfrac{1}{3}\displaystyle\int_0^1 \left(1+2x\right)^{3/2}dx=\sqrt{3}-\dfrac{1}{15}\left(1+2x\right)^{5/2}\biggr|_0^1\\ =\dfrac{2}{5}\sqrt{3}+\dfrac{1}{15}$$

An easier method, as suggested by @N.S., would be to use the substitution $u=1+2x.$ Then the integral would simply be

$\dfrac{1}{2}\displaystyle\int_1^3 \dfrac{u-1}{2}\sqrt{u}du.$