What is the integral of 1/(64+x^2) ?

The integral of 1/(64+x^2) is 1/8 arctan(x/8)+C

Popular >

integral of 1/(64+x^2)

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

\frac{1}{8} arctan (\frac{x}{8}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{8 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{8} arctan (u)

Substitute back

u = \frac{x}{8}

= \frac{1}{8} arctan (\frac{x}{8})

Add a constant to the solution

= \frac{1}{8} arctan (\frac{x}{8}) + C

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

2 x^{2} + x - 2

\int_{0}^{\infty} \frac{5}{x ^{3}} d x

0 = x^{2} - x y + 2 y - 2, at x = - 2

\frac{\partial}{\partial x} (x^{3} \cdot y^{2} + 4 x^{2} - 3 x y)

What is the integral of 1/(64+x^2) ?
The integral of 1/(64+x^2) is 1/8 arctan(x/8)+C