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

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

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integral of 1/(1+64x^2)

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

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

Solution steps

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

Apply Integral Substitution

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

Use the common integral:

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

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

Substitute back

u = 8 x

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

Add a constant to the solution

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

\int_{0}^{\infty} t e^{- t^{2}} d t

\frac{d}{d x} (\frac{x}{x - 9})

\int (3 x - 5)^{\frac{1}{2}} d x

y^{^{''}} = 6 x - 12

y = 6 x^{4} - 6 x^{2}, y = 3 x^{2}

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