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Hello I am studying for a mock test that is coming up, a question very similar to this one will be on this test, I have no idea how to complete this type of question, I have been given some vague notes about how I should turn it into a arctan function (??? I have no idea how to)

Here is the question:

$$I=\int\frac{5x^{2}-10x}{(9x^{2}-16)(x^{2}+1)}\:\mathrm{d}x$$

I've 'Simplified' down to:

$$\frac{5x^2-10x}{(3x-4)(3x-4)(x^2+1)}$$

Here is the stage I have gotten to so far:

$$I = \int\left(\frac{A}{3x-4} + \frac{B}{3x-4} + Cx+\frac{D}{x^2+1}\right)\:\mathrm{d}x$$

Thomas Russell
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Stephen
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2 Answers2

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See Partial Fraction Decomposition

$$\dfrac{5x^2-10x}{(9x^2-16)(x^2+1)}=\dfrac A{3x-4}+\dfrac B{3x+4}+\dfrac{Cx+D}{x^2+1}$$

$$\implies5x^2-10x$$ $$=A(3x+4)(x^2+1)+B(3x-4)(x^2+1)+(Cx+D)(3x+4)(3x-4)$$

Set $3x-4=0,3x+4=0$ one by one to find $A,B$

Now compare the coefficients of $x^3,$ $$0=3A+3B+9C\iff C=?$$

compare the constants $$4A-4B-16D=0\iff D=?$$

lab bhattacharjee
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Hint. You may want to use

$$ \int\frac1{a x+b}dx=\frac1a\log \left| a x+b\right|+C \quad (a\neq0)\tag1 $$

and

$$ \int\frac{x}{x^2+1}dx=\frac12\int\frac{2x}{x^2+1}dx=\frac12\int\frac{(x^2+1)'}{x^2+1}dx=\frac12\log \left( x^2+1\right)+C \tag2 $$

and

$$ \begin{align} \int \frac{1}{x^2+1}dx&=\arctan x +C \tag3 \end{align} $$

Olivier Oloa
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