# Integration by parts using u-substitution and square root

I have to use the technique of integration by parts to evaluate the integrals. I'm having trouble with a particular problem:

$$\int x (\sqrt{x+2}) dx$$

I'm using u-substitution, but since I'm also integrating by parts, my "u-substitution" will be using the variable "g"

$g=x+2$

$\frac {dg}{dx} = 1 dx = dx$

Now the original problem is:

$$\int x (\sqrt{g}) dx = \int x g^\frac 12 dx$$

So, I begin integrating by parts:

$$= (x)(\frac 23 g^\frac 32) -\int (\frac 23 g^\frac 32)(dx)$$

I take out the constant, then find the integral:

$$= (x)(\frac 23 g^\frac 32) - (\frac 23) (\frac 25 g^\frac 52) + C$$

Which then gives me:

$$= (\frac {2x}3 g^\frac 32) - (\frac 4{15} g^\frac 52) + C$$

This is where I'm stuck, and I feel that I've missed an early step in the substitution (something to do with x = u - 1?).

The answer I should eventually arrive at is:

$$= \frac 2{15} (x+2)^\frac 32 (3x-4)+C$$

• If $g = x + 2$ then $g - 2 = x$ so your integral becomes $$\int (g - 2) \sqrt{g} \ \ dg$$ not $$\int x \sqrt{g} \ \ dx$$ Feb 24, 2015 at 14:28

There really isn't any need for integration by parts.

If $g = x+2$, then $x = g-2$, and as you note, $dg = dx$. So your integral becomes $$\int (g-2)(g^{1/2}) \,dg = \int \left(g^{3/2} - 2g^{1/2}\right)\,dg$$

Can you take it from here?

• Is finding $x = g - 1$ as simple as solving for $x$ in $g = x + 2$? This part of substitution kept me awake lastnight. Feb 24, 2015 at 14:43
• Yes, but we used $g = x+2$, so to solve for $x$ we just subtract $2$ from each side of the equation to get $x = g-2.$ Feb 24, 2015 at 14:47
• Great response! I had to break this problem down into parts, so I chose the other answer. However, this answer is equally as useful, thanks! Feb 26, 2015 at 16:08

You may just perform an integration by parts: \begin{align} \int x (\sqrt{x+2}) dx&=x \:\frac 23 (x+2)^\frac 32-\frac 23\int (x+2)^\frac 32 dx\\\\ &=x \:\frac 23 (x+2)^\frac 32-\frac 23 \times\frac 25(x+2)^\frac 52 +C\\\\ &=\frac 23 x(x+2)^\frac 32-\frac 4{15}(x+2)^\frac 52 +C\\\\ &=\left(\frac 23 x-\frac 4{15}(x+2)\right)(x+2)^\frac 32 +C\\\\ &=\left(\color{red}{\frac 2{15}}\times5 x-\color{red}{\frac {2\color{black}{\times 2}}{15}}(x+2)\right)(x+2)^\frac 32 +C\\\\ &=\color{red}{\frac 2{15}}\left[\color{blue}{5x-2(x+2)}\right](x+2)^\frac 32 +C\\\\ &=\color{red}{\frac 2{15}}\left(\color{blue}{5x-2x-4}\right)(x+2)^\frac 32 +C\\\\ & =\frac 2{15} (x+2)^\frac 32 (3x-4)+C \end{align}

• Between your last step (the answer) and the one before it, I'm having a difficult time seeing what was done. Could you please elaborate on that part? Feb 24, 2015 at 14:59
• @WesFoster I've added some details, hoping to be clear. Thanks. Feb 24, 2015 at 23:20
• That is so helpful! Thanks so much! Feb 26, 2015 at 16:07