What is the integral of (x^4)/(x^2+a^2) ?

The integral of (x^4)/(x^2+a^2) is (x^3)/3-a^2x+a^3arctan(x/a)+C

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integral of (x^4)/(x^2+a^2)

\int \frac{x ^{4}}{x ^{2} + a ^{2}} d x

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

Solution steps

\int \frac{x ^{4}}{x ^{2} + a ^{2}} d x

Apply Integral Substitution

= \int \frac{a ^{3} u ^{4}}{u ^{2} + 1} d u

Take the constant out:

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

= a^{3} \cdot \int \frac{u ^{4}}{u ^{2} + 1} d u

Long division

\frac{u ^{4}}{u ^{2} + 1} : u^{2} - 1 + \frac{1}{u ^{2} + 1}

= a^{3} \cdot \int u^{2} - 1 + \frac{1}{u ^{2} + 1} 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

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

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= a^{3} (\frac{u ^{3}}{3} - u + arctan (u))

Substitute back

u = \frac{x}{a}

= a^{3} (\frac{( \frac{x}{a} ) ^{3}}{3} - \frac{x}{a} + arctan (\frac{x}{a}))

Simplify

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

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

Add a constant to the solution

= \frac{x ^{3}}{3} - a^{2} x + a^{3} arctan (\frac{x}{a}) + C

\int 4 x^{3} (2 x^{4} + 4)^{7} d x

ln^{2} (x)

\int 1 ln (x) d x

\int (x^{3} + 5 x^{2} - 2) d x

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

What is the integral of (x^4)/(x^2+a^2) ?
The integral of (x^4)/(x^2+a^2) is (x^3)/3-a^2x+a^3arctan(x/a)+C