# Why doesn't integration by parts work here?

I am in Calculus II and this is one of our exercises for integration:

$$\int x(x+3)^{-1/2} dx$$

This looked to me like a sober example of when to use Integration By Parts, so I went with that:

$$u = x$$

From which I got the $$du = dx$$

Then I took $$dv = (x+3)^{-1/2}$$

And found $$v=2\sqrt{x+3}$$ using a simple substitution.

I then proceeded with the formula $$\int{f(x)g'(x)}dx = f(x)g(x) - \int{f'(x)g(x)}dx$$

which gave me the following: $$2x\sqrt{x+3} - \int{2\sqrt{x+3}} dx$$

There is no point continuing beyond this point in my calculation as Wolfram Alpha has told me that the result of the above computation is different from the initial integral.

I have starred at this for a while. I had a friend stare at it too and neither of us could see a mistake.

This is my first time using LaTeX so I may have done a typo somewhere. Sorry about that.

Here are the two outputs:

Me: $$2x\sqrt{x+3}-4/3(x+3)^{3/2} + C$$ Alpha: $$2/3(x-6)\sqrt{x+3} + C$$

• Remember that these are indefinite integrals, and your answer may disagree by a constant from Alpha (and still be correct). – vadim123 Mar 20 '18 at 0:59
• The answer I get from evaluating your last expression (after some algebraic manipulation) gives the same result that WolframAlpha spits out. – Mauve Mar 20 '18 at 0:59
• Integration by parts should work here. What do you get as your solution? – BSplitter Mar 20 '18 at 1:01
• If you are unsure if you solution is the same as the solution that wolfram alpha has, try plotting both functions in Desmos. (desmos.com. ) – BSplitter Mar 20 '18 at 1:03
• Will think of comparing them more next time before posting. Thanks everybody! – Slava Knyazev Mar 20 '18 at 1:06

WolframAlpha says the original integral comes to $\frac{2}{3}(x - 6)\sqrt{x + 3} + c$. It says the new integral (after integration by parts) is $2x\sqrt{x + 3} - \frac{4}{3}(x + 3)^{3/2} + c$.

If we factor $\sqrt{x + 3}$ out of the second one, we have $(2x - \frac{4}{3}(x + 3))\sqrt{x + 3} + c$. $2x - \frac{4}{3}(x + 3) = 2x - \frac{4}{3}x - 4 = \frac{2}{3}x - 4 = \frac{2}{3}(x - 6)$. So these two expressions are the same!

Here's a tip: If you find yourself in a situation where you think you have two different answers, try plugging in several values of $x$. There are three possibilities:

• You get the same value for each no matter which $x$ you use (like this situation); that means they're secretly the same, and you should look for an algebraic way to simplify one into the other.
• You get different values, but the two expressions always differ by the same amount; that means they're just getting different $+c$ values.
• You get wildly different values; that means you did something wrong!
• +1 for the great general tips next to answering the question! – Jan Mar 20 '18 at 13:19

You have $$2x\sqrt{x+3} - \int{2\sqrt{x+3}} dx$$

For the integral $$\int{2\sqrt{x+3}} dx$$

Make the substitution $$x+3=u$$ and you have $$\int{2\sqrt{x+3}} dx =2\int{\sqrt{u}} du$$

You know how to finish it.

Without any substitution and without any integration by parts:

$$\int \frac{x}{\sqrt{x+3}}\, dx= \int \frac{x+3-3}{\sqrt{x+3}}\, dx= \int \sqrt{x+3}\, dx - 3\int \frac{1}{\sqrt{x+3}}\, dx =$$ $$= \frac{2}{3}(x+3)\sqrt{x+3}-6\sqrt{x+3} =\frac{2}{3}(x-6)\sqrt{x+3}$$

I believe you can go further in your computation by assuming $u = x + 3$ ; you'll eventually end up at the same result as provided by Wolfram Alpha. In my solution here, I would like to present another plausible solution for the given integral, which however does not use Integration by parts.

Let $u = x + 3 \implies x = u - 3$ and $du = dx$

Therefore the integral reduces to :- $$\int{\frac{u - 3}{\sqrt{u}}} du$$

$$= \int{[\sqrt{u} - 3u^{\frac{-1}{2}}}] du$$ $$= \frac 23 (x+3)^{\frac 32} - 6\sqrt{x+3} + C$$ $$= \sqrt{x+3} * [\frac23 (x+3) - 6]$$ which on solving a little further (Solve it yourself to confirm) yields the following result -

$$= \frac23 \sqrt{x+3} (x - 6) + C$$

as given by Wolfram Alpha as well.