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

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

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

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

\frac{1}{2} (ln a^{2} + x^{2} + \frac{a ^{2}}{a ^{2} + x ^{2}}) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

Substitute back

u = a^{2} + x^{2}

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

Simplify

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

Add a constant to the solution

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

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

\int_{- 2}^{4} \frac{1}{2} d x

\frac{d}{d x} (e^{- 2 x} cos (4 x))

\frac{d}{d x} (c e^{x})

f (x) = 5 x^{2} - x^{3}, at x = 1

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