What is the integral of (39x^2)/((x+1)(x^2+1)) ?

The integral of (39x^2)/((x+1)(x^2+1)) is 39(1/2 ln|x+1|+1/2 (1/2 ln|x^2+1|-arctan(x)))+C

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integral of (39x^2)/((x+1)(x^2+1))

\int \frac{39 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

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

Solution steps

\int \frac{39 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

Take the constant out:

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

= 39 \cdot \int \frac{x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

Take the partial fraction of

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

= 39 \cdot \int \frac{1}{2 ( x + 1 )} + \frac{x - 1}{2 ( 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

= 39 (\int \frac{1}{2 ( x + 1 )} d x + \int \frac{x - 1}{2 ( x ^{2} + 1 )} d x)

\int \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣

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

= 39 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x)))

Add a constant to the solution

= 39 (\frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 - arctan (x))) + C

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What is the integral of (39x^2)/((x+1)(x^2+1)) ?
The integral of (39x^2)/((x+1)(x^2+1)) is 39(1/2 ln|x+1|+1/2 (1/2 ln|x^2+1|-arctan(x)))+C