$$\frac{1}{10} \int \frac{3x+20}{x^2+10}dx$$ I was trying to factor out the denominator: $$\frac{1}{10} \int \frac{3x+20}{x^2+(\sqrt{10})^2}dx$$
$$\frac{1}{10} \int \frac{3x+20}{(x+\sqrt{10})^2-2x\sqrt{10}}dx$$
I was trying to factor the denominator to use partial fractions but failed how should I approach this integral?
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$${\frac{3x+20}{x^2+10}=\frac{\frac{3}2+\sqrt{10}i}{x+\sqrt{10}i}+\frac{\frac{3}2-\sqrt{10}i}{x-\sqrt{10}i}}$$
Lets integrate,
$$\int \frac{3x+20}{x^2+10}{=\int \frac{\frac{3}2+\sqrt{10}i}{x+\sqrt{10}i}+\int \frac{\frac{3}2-\sqrt{10}i}{x-\sqrt{10}i}\\ =\left(\frac{3}2+\sqrt{10}i\right)\ln\left(x+\sqrt{10}i\right)+\left(\frac{3}2-\sqrt{10}i\right)\ln\left(x-\sqrt{10}i\right)+c}$$
Now, $$\ln\left(x+\sqrt{10}i\right)=\ln\left(\sqrt{x^2+10}\,e^{i\arctan{\frac{\sqrt{10}}{x}}}\right)=\ln\sqrt{x^2+10} +{i\arctan{\frac{\sqrt{10}}{x}}}\\ \ln\left(x-\sqrt{10}i\right)=\ln\left(\sqrt{x^2+10}\,e^{i\arctan{\frac{-\sqrt{10}}{x}}}\right)=\ln\sqrt{x^2+10} -{i\arctan{\frac{\sqrt{10}}{x}}}\\$$
Puting the values, $$\int \frac{3x+20}{x^2+10}=3\ln\sqrt{x^2+10}-2\sqrt{10}\arctan\frac{\sqrt{10}}{x}+c$$
This is how to "factor the denominator to use partial fractions."
The idea is to split it into two integrals, one of the form $\int f'/f\,dx$, the other $\int dx/(x^2+a^2)$, which equals $\tfrac1a\tan^{-1}\tfrac xa$: \begin{align*} \int\frac{3x+20}{x^2+10}\,dx&=\frac32\int\frac{2x}{x^2+10}\,dx+20\int\frac{dx}{x^2+(\sqrt{10})^2}\\[15pt] &=\frac32\log(x^2+10)+\frac{20}{\sqrt{10}}\tan^{-1}\Big(\frac x{\sqrt{10}}\Big). \end{align*}