What is the integral of (x^2)/(x^2+81) ?

The integral of (x^2)/(x^2+81) is -9arctan(x/9)+x+C

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integral of (x^2)/(x^2+81)

\int \frac{x ^{2}}{x ^{2} + 81} d x

- 9 arctan (\frac{x}{9}) + x + C

Solution steps

\int \frac{x ^{2}}{x ^{2} + 81} d x

\frac{x ^{2}}{x ^{2} + 81} = - \frac{81}{x ^{2} + 81} + 1

= \int - \frac{81}{x ^{2} + 81} + 1 d x

Apply Integral Substitution

= \int 9 (- \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

= 9 \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

= 9 (- \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

= 9 (- arctan (u) + u)

Substitute back

u = \frac{x}{9}

= 9 (- arctan (\frac{x}{9}) + \frac{x}{9})

Simplify

9 (- arctan (\frac{x}{9}) + \frac{x}{9}) : - 9 arctan (\frac{x}{9}) + x

= - 9 arctan (\frac{x}{9}) + x

Add a constant to the solution

= - 9 arctan (\frac{x}{9}) + x + C

\frac{y ^{^{'}} - e ^{- t} + 5}{y} = - 5, y (0) = - 4

(5, - 1), (- 3, 4)

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

\int 4 y^{4} d y

x = y (2 - y), x = 0

What is the integral of (x^2)/(x^2+81) ?
The integral of (x^2)/(x^2+81) is -9arctan(x/9)+x+C