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

The integral of (x^4)/(x^2+4) is 8((x^3)/(24)-x/2+arctan(x/2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

8 (\frac{x ^{3}}{24} - \frac{x}{2} + arctan (\frac{x}{2})) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{8 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

= 8 \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}

= 8 \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

= 8 (\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)

= 8 (\frac{u ^{3}}{3} - u + arctan (u))

Substitute back

u = \frac{x}{2}

= 8 (\frac{( \frac{x}{2} ) ^{3}}{3} - \frac{x}{2} + arctan (\frac{x}{2}))

Simplify

8 (\frac{( \frac{x}{2} ) ^{3}}{3} - \frac{x}{2} + arctan (\frac{x}{2})) : 8 (\frac{x ^{3}}{24} - \frac{x}{2} + arctan (\frac{x}{2}))

= 8 (\frac{x ^{3}}{24} - \frac{x}{2} + arctan (\frac{x}{2}))

Add a constant to the solution

= 8 (\frac{x ^{3}}{24} - \frac{x}{2} + arctan (\frac{x}{2})) + C

\int tan^{3} (3 x) sec^{2} (3 x) d x

\int x csc^{2} (x^{2}) d x

\int \frac{1}{2 + 5 cos ( x )} d x

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

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

What is the integral of (x^4)/(x^2+4) ?
The integral of (x^4)/(x^2+4) is 8((x^3)/(24)-x/2+arctan(x/2))+C