What is the integral of 1/((x^2+1)(x-3)^2) ?

The integral of 1/((x^2+1)(x-3)^2) is 1/50 (3/2 ln|x^2+1|+4arctan(x))-3/50 ln|x-3|-1/(10(x-3))+C

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integral of 1/((x^2+1)(x-3)^2)

\int \frac{1}{( x ^{2} + 1 ) ( x - 3 ) ^{2}} d x

\frac{1}{50} (\frac{3}{2} ln x^{2} + 1 + 4 arctan (x)) - \frac{3}{50} ln ∣ x - 3 ∣ - \frac{1}{10 ( x - 3 )} + C

Solution steps

\int \frac{1}{( x ^{2} + 1 ) ( x - 3 ) ^{2}} d x

Take the partial fraction of

\frac{1}{( x ^{2} + 1 ) ( x - 3 ) ^{2}} : \frac{3 x + 4}{50 ( x ^{2} + 1 )} - \frac{3}{50 ( x - 3 )} + \frac{1}{10 ( x - 3 ) ^{2}}

= \int \frac{3 x + 4}{50 ( x ^{2} + 1 )} - \frac{3}{50 ( x - 3 )} + \frac{1}{10 ( x - 3 ) ^{2}} 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{3 x + 4}{50 ( x ^{2} + 1 )} d x - \int \frac{3}{50 ( x - 3 )} d x + \int \frac{1}{10 ( x - 3 ) ^{2}} d x

\int \frac{3 x + 4}{50 ( x ^{2} + 1 )} d x = \frac{1}{50} (\frac{3}{2} ln x^{2} + 1 + 4 arctan (x))

\int \frac{3}{50 ( x - 3 )} d x = \frac{3}{50} ln ∣ x - 3 ∣

\int \frac{1}{10 ( x - 3 ) ^{2}} d x = - \frac{1}{10 ( x - 3 )}

= \frac{1}{50} (\frac{3}{2} ln x^{2} + 1 + 4 arctan (x)) - \frac{3}{50} ln ∣ x - 3 ∣ - \frac{1}{10 ( x - 3 )}

Add a constant to the solution

= \frac{1}{50} (\frac{3}{2} ln x^{2} + 1 + 4 arctan (x)) - \frac{3}{50} ln ∣ x - 3 ∣ - \frac{1}{10 ( x - 3 )} + C

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What is the integral of 1/((x^2+1)(x-3)^2) ?
The integral of 1/((x^2+1)(x-3)^2) is 1/50 (3/2 ln|x^2+1|+4arctan(x))-3/50 ln|x-3|-1/(10(x-3))+C