integration by parts

Question

!!!! PD: I did a little change in the denominator !!!!

I need to solve this integral using integration by parts.

$\displaystyle\int\frac{x\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}$

Thanks!

PS: I know that I can to do:

$\displaystyle\int\frac{x\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}=\int\frac{(x-c)\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}+c\int\frac{dx}{\sqrt{(a^2+b^2)+(x-c)^2}}$

but according to the book it is easier using integration by part.

Using the idea in "PS" is easy using trig substitution, but I need use integration by parts. According to the book is the easier way. — yemino, Sep 20 '13 at 20:52
Consider the possibility that the book is wrong. Not that it necessarily means much, but I don't see how integration by parts would make it easier. — Daniel Fischer, Sep 20 '13 at 21:01
What does "PD" mean? The change in the denominator doesn't affect the problem. Whether it's $a^2+b^2$ or $(a^2+b^2)^2$, it's just a positive number, which you could call $k^2$ if it makes your life easier. (I guess you want to handle $a=b=0$ separately.) — Jonas Meyer, Sep 20 '13 at 21:19
PD = P.D.: post data (just to emphasize that the change was after sending the message). — yemino, Sep 20 '13 at 21:23

score 1 · Answer 1 · answered Sep 19 '22 at 03:18

Let $x-c=\sqrt{a^2+b^2} \tan \theta,$ then $d x=\sqrt{a^2+b^2} \sec ^2 \theta d \theta$ and $$ \begin{aligned} \int \frac{x d x}{\sqrt{\left(a^2+b^2\right)+(x-c)^2}} &=\int \frac{c+\sqrt{a^2+b^2 \tan \theta}}{\sqrt{a^2+b^2 \sec \theta}} \sqrt{a^2+b^2 \sec ^2 \theta} d \theta\\ &=\int\left(c+\sqrt{a^2+b^2} \tan \theta\right) \sec \theta d \theta \\ &=c \ln |\sec \theta+\tan \theta|+\sqrt{a^2+b^2} \sec \theta+C\\ &=c \ln \left|\frac{x-c+\sqrt{a^2+b^2+(x-c)^2}}{\sqrt{a^2+b^2}}\right|+\sqrt{a^2+b^2+(x-c)^2}+C \end{aligned} $$

score 0 · Answer 2 · answered Sep 19 '22 at 03:58

You don't actually need integration by parts. First, observe the following:

$$\left[\sqrt{a^2+b^2+(x-c)^2}\right]'=\frac {x-c}{\sqrt{a^2+b^2+(x-c)^2}}$$

Integrating both sides and isolating the integral under question gives

$$\int\frac x{\sqrt{a^2+b^2+(x-c)^2}}\,\mathrm dx=\sqrt{a^2+b^2+(x-c)^2}+c\int\frac {\mathrm dx}{\sqrt{a^2+b^2+(x-c)^2}}$$

The latter integral can be evaluated using an Euler Substitution. A more detailed explanation of how to perform an Euler Substitution can be found on the Wikipedia page, but the substitution to make here is

$$t=x-c+\sqrt{a^2+b^2+(x-c)^2}$$

Differentiating gives

\begin{align*} \mathrm dt & =1+\frac {x-c}{\sqrt{a^2+b^2+(x-c)^2}}\,\mathrm dx\\ & =\frac {x-c+\sqrt{a^2+b^2+(x-c)^2}}{\sqrt{a^2+b^2+(x-c)^2}}\,\mathrm dx\\ & =\frac t{\sqrt{a^2+b^2+(x-c)^2}}\,\mathrm dx \end{align*}

As in

$$\frac {\mathrm dt}t=\frac {\mathrm dx}{\sqrt{a^2+b^2+(x-c)^2}}$$

The integral is then

\begin{align*} & \int\frac x{\sqrt{a^2+b^2+(x-c)^2}}\,\mathrm dx\\ & =\sqrt{a^2+b^2+(x-c)^2}+c\int\frac {\mathrm dt}t\\ & =\sqrt{a^2+b^2+(x-c)^2}+c\log\left(x-c+\sqrt{a^2+b^2+(x-c)^2}\right)+K \end{align*}

integration by parts

2 Answers2