What is the integral of (x^2)/(x^2+64) ?

The integral of (x^2)/(x^2+64) is -8arctan(x/8)+x+C

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integral of (x^2)/(x^2+64)

\int \frac{x ^{2}}{x ^{2} + 64} d x

- 8 arctan (\frac{x}{8}) + x + C

Solution steps

\int \frac{x ^{2}}{x ^{2} + 64} d x

\frac{x ^{2}}{x ^{2} + 64} = - \frac{64}{x ^{2} + 64} + 1

= \int - \frac{64}{x ^{2} + 64} + 1 d x

Apply Integral Substitution

= \int 8 (- \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

= 8 \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

= 8 (- \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

= 8 (- arctan (u) + u)

Substitute back

u = \frac{x}{8}

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

Simplify

8 (- arctan (\frac{x}{8}) + \frac{x}{8}) : - 8 arctan (\frac{x}{8}) + x

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

Add a constant to the solution

= - 8 arctan (\frac{x}{8}) + x + C

\frac{d}{d x} (10 arccot (x))

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

\frac{d}{d x} (x y^{2} + x^{2} y^{3} + x^{3} y^{4})

x^{2} + y^{2} = 25, (- 3, 4)

\frac{d}{d x} (x^{3} (1 - x)^{2})

What is the integral of (x^2)/(x^2+64) ?
The integral of (x^2)/(x^2+64) is -8arctan(x/8)+x+C