What is the integral of (x^2)/(x^2+5) ?

The integral of (x^2)/(x^2+5) is -sqrt(5)arctan(x/(sqrt(5)))+x+C

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integral of (x^2)/(x^2+5)

integral

\frac{x ^{2}}{x ^{2} + 5}

- 5 arctan (\frac{x}{5}) + x + C

Solution steps

\int \frac{x ^{2}}{x ^{2} + 5} d x

\frac{x ^{2}}{x ^{2} + 5} = - \frac{5}{x ^{2} + 5} + 1

= \int - \frac{5}{x ^{2} + 5} + 1 d x

Apply Integral Substitution

= \int 5 (- \frac{1}{u ^{2} + 1} + 1) d u

Take the constant out:

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

= 5 \cdot \int - \frac{1}{u ^{2} + 1} + 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

= 5 (- \int \frac{1}{u ^{2} + 1} d u + \int 1 d u)

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

\int 1 d u = u

= 5 (- arctan (u) + u)

Substitute back

u = \frac{x}{5}

= 5 (- arctan (\frac{x}{5}) + \frac{x}{5})

Simplify

5 (- arctan (\frac{x}{5}) + \frac{x}{5}) : - 5 arctan (\frac{x}{5}) + x

= - 5 arctan (\frac{x}{5}) + x

Add a constant to the solution

= - 5 arctan (\frac{x}{5}) + x + C

in t e g r a l \frac{( e ^{x} ) ^{2}}{x}

in t e g r a l - x^{2} ln (x)

in t e g r a l \frac{x ^{2}}{x ^{2} + 1}

in t e g r a l y (4 - y)^{\frac{1}{2}}

in t e g r a l (\frac{e}{2})^{x}

What is the integral of (x^2)/(x^2+5) ?
The integral of (x^2)/(x^2+5) is -sqrt(5)arctan(x/(sqrt(5)))+x+C