0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I cannot make sense of your first equation. $\endgroup$ – Andrew Chin Oct 27 at 19:22
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @Sebastiano What is $\int v du$? Remember thet $v= \int \sqrt{1+2x} dx$. $\endgroup$ – N. S. Oct 27 at 22:18
  • $\begingroup$ I'm sorry. I have not seen :-( I will remove my comments. $\endgroup$ – Sebastiano Oct 28 at 8:07
0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ okay, but I do not want to use the substitution method, but rather learn how to do integration by parts properly $\endgroup$ – fazan Oct 27 at 19:39
  • $\begingroup$ okay so the general formula can be derived directly from the product rule. it is $\displaystyle\int udv = uv - \displaystyle\int vdu$ $\endgroup$ – Simon Fraser Oct 27 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.