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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!

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My advice is that if you are going to need to do partial fractions, you might as well do it from the beginning, rather than duplicate your effort later. That is to say, try from the start to write $$\frac{2x-3}{x^3+10x} = \frac{Ax+B}{x^2+10} + \frac{C}{x}.$$ Using the cover-up method, we get $$x(Ax+B) + (x^2+10)C = 2x-3,$$ hence setting $x = 0$ yields $C = -3/10$. Then comparing the remaining coefficients gives $(A+C)x^2 + Bx + 10C = 2x-3$, or $A = 3/10$, and $B = 2$. So the desired decomposition is $$\frac{2x-3}{x^3+10x} = \frac{1}{10} \left( \frac{3x+20}{x^2+10} - \frac{3}{x} \right),$$ which is more conveniently written $$\frac{3}{20} \cdot \frac{2x}{x^2 + 10} + 2 \cdot \frac{1}{x^2+10} - \frac{3}{10} \cdot \frac{1}{x}.$$ The first term integrates to a logarithm; the second an inverse tangent, and the last is another logarithm. The result is more simplified than the textbook's answer: $$\frac{3}{20} \log |x^2+10| + \frac{2}{\sqrt{10}} \tan^{-1} \frac{x}{\sqrt{10}} - \frac{3}{10} \log |x| + C,$$ or equivalently, $$\frac{3}{10} \log \left| \frac{\sqrt{x^2+10}}{x} \right| + \frac{2}{\sqrt{10}} \tan^{-1} \frac{x}{\sqrt{10}} + C.$$

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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*}

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  • $\begingroup$ Thank you that clears things up. Why do I need to use Bx+C instead of simply B? $\endgroup$ Apr 16 '15 at 4:26
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    $\begingroup$ @Quantaliinuxite since the denominator is $x^2+10$, which is a second degree polynomial so the numerator can be of degree at most $1$. That is why we use $Bx+C$ instead of just $B$ (which will be degree $0$). $\endgroup$
    – Anurag A
    Apr 16 '15 at 4:28
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You neglected the chain rule: $$ \int \frac{dx}{\left(\frac x {\sqrt{10}}\right)^2 + 1} = \sqrt{10}\arctan\left(\frac x {\sqrt{10}} \right) + C $$

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