Question

$\int \frac{4x^5-7x^4+8x^3-2x^2+4x-7}{x^2(x^2+1{)}^2}dx$.

Answer 1

$I=\int \frac{4x^5–7x^4+8x^3–2x^2+4x-7}{x^2(x^2+1{)}^2}dx$

We can split it into partial fractions as $\frac{A}{x^2}+\frac{B}{x}+\frac{Cx+D}{x^2+1}+\frac{Ex+F}{(x^2+1{)}^2}.....\left(1\right)$

Multiplying both sides by $x^2$ and putting $x=0$ we get $A=-7$

Similarly multiplying by $(x^2+1{)}^2$ and substituting $x=i$ we get

$Ei+F=12\therefore E=0,F=12$

Since there are repeating factors $B,C$ and $D$ which can not be found by this method.

to solve it we use the substitution.

This is done by substituting different and convenient values for $x.$

$x=1$ on either side gave $2B+C+D=8\dots \dots \left(2\right)$

$x=-1$ gives $-2B-C+D=-8\dots \dots .\left(3\right)$

From these 2 relations we get $D=0$

Next $x=2$ gives $5B+4C=20\dots \dots \left(4\right)$

From $\left(2\right)$ and $\left(4\right)$ we get $B=4$ and $C=0$

$\therefore I=\int \frac{-7}{x^2}+\frac{4}{x}+\frac{0\times x+0}{x^2+1}+\frac{0\times x+12}{(x^2+1{)}^2}dx$

$=\int \frac{-7}{x^2}+\frac{4}{x}+\frac{12}{(x^2+1{)}^2}dx$

First two are direct

For third put $x=tan\theta \Rightarrow dx=se{c}^2\theta d\theta$

$I=\frac{14}{x}+4log|x|+12\int \frac{se{c}^2\theta }{se{c}^4\theta }d\theta$

$I=\frac{14}{x}+4log|x|+12\int co{s}^2\theta d\theta$

$\int co{s}^2\theta d\theta =\int \frac{1+cos2\theta }{2}d\theta =\frac{\theta }{2}+\frac{sin2\theta }{4}=\frac{ta{n}^{-1}x}{2}+\frac{2x}{1+x^2}\times \frac{1}{4}$

$\therefore I=\frac{14}{x}+4log|x|+6ta{n}^{-1}x+\frac{6x}{1+x^2}$