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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

\frac{d y}{d x} = e^{3.1 x - 3 y}

\frac{\partial}{\partial x} (x^{4} sin (2 t))

sin (x) + \frac{4}{3} cot (x)

\int \frac{1}{x ^{2} - 7 x + 12} d x

x y^{2} \frac{d y}{d x} = y^{3} - 3 x^{3}, y (1) = 5

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