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

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

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

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

ln x^{2} + 1 + \frac{1}{x ^{2} + 1} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u - 1}{2 u ^{2}} d u

= 2 \cdot \int \frac{u - 1}{2 u ^{2}} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{u - 1}{u ^{2}} d u

Expand

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{u} - \frac{1}{u ^{2}} d u

Apply the Sum Rule:

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

= 2 \cdot \frac{1}{2} (\int \frac{1}{u} d u - \int \frac{1}{u ^{2}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= 2 \cdot \frac{1}{2} (ln ∣ u ∣ - (- \frac{1}{u}))

Substitute back

u = x^{2} + 1

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

Simplify

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

= ln x^{2} + 1 + \frac{1}{x ^{2} + 1}

Add a constant to the solution

= ln x^{2} + 1 + \frac{1}{x ^{2} + 1} + C

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

\frac{2 x}{x ^{2} + 1}

x^{3}, at x = 0

(0, - 5), (- 3, 5)

x \to 0 - lim (e^{\frac{4}{x}})

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