# Calculation of integral

• MHB
• mathmari

#### mathmari

Gold Member
MHB
Hey!

I want to calculate the integral $$\int\frac{1}{(x+4)(x^2-8x+19)}\, dx$$ I have done the following : $$\frac{1}{(x+4)(x^2-8x+19)}=\frac{1}{67}\frac{1}{x+4}+\frac{1}{67}\frac{12-x}{x^2-8x+19}$$ and so we get \begin{align*}\int\frac{1}{(x+4)(x^2-8x+19)}\, dx&=\frac{1}{67}\int \frac{1}{x+4}\, dx+\frac{1}{67}\int \frac{12-x}{x^2-8x+19}\, dx\\ & =\frac{1}{67}\cdot \ln |x+4|+\frac{1}{67}\int \frac{8+4-x}{x^2-8x+19}\, dx\\ & =\frac{1}{67}\cdot \ln |x+4|+\frac{1}{67}\int \frac{8 }{x^2-8x+19}\, dx+\frac{1}{67}\int \frac{4-x}{x^2-8x+19}\, dx \\ & =\frac{1}{67}\cdot \ln |x+4|+\frac{1}{67}\int \frac{8 }{x^2-8x+19}\, dx +\frac{1}{67}\cdot \ln | x^2-8x+19| \end{align*} Is everything correct so far? How could we continue to calculate the middle term? :unsure:

Hey mathmari!

If I take the derivative of the last term with "ln" in it, I do not get what we integrated. (Worried)

To integrate the middle term, we can use the derivative of arctan. What is that derivative? (Wondering)

If I take the derivative of the last term with "ln" in it, I do not get what we integrated. (Worried)

Ah I forgot to divide by $-2$. It should be:
\begin{align*}\int\frac{1}{(x+4)(x^2-8x+19)}\, dx&=\frac{1}{67}\int \frac{1}{x+4}\, dx+\frac{1}{67}\int \frac{12-x}{x^2-8x+19}\, dx\\ & =\frac{1}{67}\cdot \ln |x+4|+\frac{1}{67}\int \frac{8+4-x}{x^2-8x+19}\, dx\\ & =\frac{1}{67}\cdot \ln |x+4|+\frac{1}{67}\int \frac{8 }{x^2-8x+19}\, dx+\frac{1}{67}\int \frac{4-x}{x^2-8x+19}\, dx\\ & =\frac{1}{67}\cdot \ln |x+4|+\frac{1}{67}\int \frac{8 }{x^2-8x+19}\, dx-\frac{1}{134}\int \frac{2x-8}{x^2-8x+19}\, dx \\ & =\frac{1}{67}\cdot \ln |x+4|+\frac{1}{67}\int \frac{8 }{x^2-8x+19}\, dx -\frac{1}{134}\cdot \ln | x^2-8x+19| \end{align*}

:unsure:

To integrate the middle term, we can use the derivative of arctan. What is that derivative? (Wondering)

It holds that $\left (\text{arctan}(x)\right )'=\frac{1}{1+x^2}$.
$$\int \frac{8 }{x^2-8x+19}\, dx=\int \frac{8 }{(x-4)^2+3}\, dx=\frac{8}{3}\cdot \int \frac{1 }{\left (\frac{x-4}{\sqrt{3}}\right )^2+1}\, dx\overset{u=\frac{x-4}{\sqrt{3}}}{=} \frac{8}{3}\cdot \int \frac{1 }{\left (u\right )^2+1}\, \sqrt{3}du=\frac{8}{\sqrt{3}}\text{arctan}(u)=\frac{8}{\sqrt{3}}\cdot \text{arctan}\left (\frac{x-4}{\sqrt{3}}\right )$$

Therefore we get \begin{align*}\int\frac{1}{(x+4)(x^2-8x+19)}\, dx& =\frac{1}{67}\cdot \ln |x+4|+\frac{1}{67}\int \frac{8 }{x^2-8x+19}\, dx -\frac{1}{134}\cdot \ln | x^2-8x+19| \\ & =\frac{1}{67}\cdot \ln |x+4|+\frac{1}{67}\cdot \frac{8}{\sqrt{3}}\cdot \text{arctan}\left (\frac{x-4}{\sqrt{3}}\right ) -\frac{1}{134}\cdot \ln | x^2-8x+19|\\ & =\frac{1}{67}\cdot \ln |x+4|+\frac{8}{67\sqrt{3}}\cdot \text{arctan}\left (\frac{x-4}{\sqrt{3}}\right ) -\frac{1}{134}\cdot \ln | x^2-8x+19| \end{align*}
Is everything correct? :unsure:

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Is everything correct?
It looks correct to me. (Nod)

It looks correct to me. (Nod)

Great! Thank you! (Sun)

Ah, one more thing... there should be a $+C$ at the end. (Blush)
Every indeterminate integral must have an indeterminate integration constant.

Ah, one more thing... there should be a $+C$ at the end. (Blush)
Every indeterminate integral must have an indeterminate integration constant.

Ahh yes (Blush)