What is the integral of (x^2)/((x-2)(x^2+1)) ?

The integral of (x^2)/((x-2)(x^2+1)) is 4/5 ln|x-2|+1/5 (1/2 ln|x^2+1|+2arctan(x))+C

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integral of (x^2)/((x-2)(x^2+1))

integral

\frac{x ^{2}}{( x - 2 ) ( x ^{2} + 1 )}

\frac{4}{5} ln ∣ x - 2 ∣ + \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x)) + C

Solution steps

\int \frac{x ^{2}}{( x - 2 ) ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{x ^{2}}{( x - 2 ) ( x ^{2} + 1 )} : \frac{4}{5 ( x - 2 )} + \frac{x + 2}{5 ( x ^{2} + 1 )}

= \int \frac{4}{5 ( x - 2 )} + \frac{x + 2}{5 ( x ^{2} + 1 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{4}{5 ( x - 2 )} d x + \int \frac{x + 2}{5 ( x ^{2} + 1 )} d x

\int \frac{4}{5 ( x - 2 )} d x = \frac{4}{5} ln ∣ x - 2 ∣

\int \frac{x + 2}{5 ( x ^{2} + 1 )} d x = \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x))

= \frac{4}{5} ln ∣ x - 2 ∣ + \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x))

Add a constant to the solution

= \frac{4}{5} ln ∣ x - 2 ∣ + \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x)) + C

in t e g r a l y = \frac{x ^{2} - cos ( x ) + 1}{x}

in t e g r a l \frac{0.001}{( 0.001 x + 1 ) ^{2}}

in t e g r a l x - 3

in t e g r a l \frac{( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )}

in t e g r a l 1. 3^{x}

What is the integral of (x^2)/((x-2)(x^2+1)) ?
The integral of (x^2)/((x-2)(x^2+1)) is 4/5 ln|x-2|+1/5 (1/2 ln|x^2+1|+2arctan(x))+C