What is the integral of (5x^2)/((x-2)(x^2+1)) ?

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

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integral of (5x^2)/((x-2)(x^2+1))

integral

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \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 )}

= 5 \cdot \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

= 5 (\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))

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

Simplify

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

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

Add a constant to the solution

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

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

9 cos^{2} (x)

x lo g_{10} (x^{2} + 5)

f (x) = (x^{2} + x + 1)^{\frac{1}{2}}

\frac{( x ^{3} - 1 )}{( x - 1 )}

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