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

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

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

integral

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

\int 1 d u = u

\int \frac{a ^{2}}{u} d u = a^{2} ln ∣ u ∣

= \frac{1}{2} (u - a^{2} ln ∣ u ∣)

Substitute back

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

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

Add a constant to the solution

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

\frac{e ^{- x}}{10}

\frac{( a ^{2} - x ^{2} )}{x}

2 y sin (x) - 4

\frac{1}{( v ^{2} + 3 )}

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

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