Where did I go wrong when doing this integral? This is my integral
$$
\int \frac{ (2x-3)}{(x^3 +10x)}\cdot dx \\
$$
This is my work
$$
\int\frac{2x}{x^3 +10x}\cdot dx-\int\frac{3}{x^3 +10x}\cdot dx\\
$$
Looking at them separately:
$$
\int\frac{2x}{x^3 +10x}\cdot dx \\= 2\int\frac{dx}{x^2 +10}
\\=\frac{2}{10}\int\frac{dx}{(\frac{x}{\sqrt{10}})^2+1}
\\=\frac{2}{10}\arctan(\frac{x}{\sqrt{10}})
$$
and
$$
\int\frac{3}{x^3 +10x}\cdot dx\\
\text{Using partial fractions to split the integrand yields}\\
\frac{3}{10}\int\frac{1}{x}\cdot dx -\frac{3}{\sqrt{10}}\int\frac{1}{x^2+10}\\
=\frac{3}{10}\ln{x} - \frac{3}{10\sqrt{10}}\arctan(\frac{x}{\sqrt{10}}) 
$$
As my final integral I therefore have:
$$
\frac{2}{10}\arctan(\frac{x}{\sqrt{10}})-\frac{3}{10}\ln{x} + \frac{3}{10\sqrt{10}}\arctan(\frac{x}{\sqrt{10}}) 
$$
However my textbook gives a radically different answer:
$$
\frac{20\arctan \left(\frac{x}{\sqrt{10}}\right)-3\cdot \sqrt{10}\left(\ln \left(\frac{x}{\sqrt{10}}\right)-\frac{\ln \left(\frac{x^2}{10}+1\right)}{2}\right)}{10\cdot \sqrt{10}}
$$
Could someone tell me where I went wrong? I can't figure it out for the life of me.
Thanks!

Your partial fraction is incorrect. Use
$$\frac{3}{x(x^2+10)}=\frac{A}{x}+\frac{Bx+C}{x^2+10}$$
to get
$$A=\frac{3}{10}, B= -\frac{3}{10} , C=0.$$
Thus
\begin{align*}
\int\frac{3}{x^3 +10x} \, dx & =\frac{3}{10}\int \frac{1}{x} \,dx - \frac{3}{10}\int \frac{x}{x^2+10} \,dx \\
& = \frac{3}{10} \ln |x| - \frac{3}{20} \ln(x^2+10) + C.
\end{align*}
