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

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) $$\dfrac{1+6\sqrt{3}}{15}$$.

• I cannot make sense of your first equation. – Andrew Chin Oct 27 at 19:22

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.

• @Sebastiano What is $\int v du$? Remember thet $v= \int \sqrt{1+2x} dx$. – N. S. Oct 27 at 22:18
• I'm sorry. I have not seen :-( I will remove my comments. – Sebastiano Oct 28 at 8:07

Let $$u=x$$ and $$dv=\sqrt{1+2x}.$$
$$x\dfrac{1}{3}\left(1+2x\right)^{3/2}\biggr|_0^1-\dfrac{1}{3}\displaystyle\int_0^1 \left(1+2x\right)^{3/2}dx=\sqrt{3}-\dfrac{1}{15}\left(1+2x\right)^{5/2}\biggr|_0^1\\ =\dfrac{2}{5}\sqrt{3}+\dfrac{1}{15}$$
An easier method, as suggested by @N.S., would be to use the substitution $$u=1+2x.$$ Then the integral would simply be
$$\dfrac{1}{2}\displaystyle\int_1^3 \dfrac{u-1}{2}\sqrt{u}du.$$
• okay so the general formula can be derived directly from the product rule. it is $\displaystyle\int udv = uv - \displaystyle\int vdu$ – Simon Fraser Oct 27 at 19:40