What is the integral of 1/(1+16x^2) ?

The integral of 1/(1+16x^2) is 1/4 arctan(4x)+C

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integral of 1/(1+16x^2)

\int \frac{1}{1 + 16 x ^{2}} d x

\frac{1}{4} arctan (4 x) + C

Solution steps

\int \frac{1}{1 + 16 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{4 ( 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}{4} \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}{4} arctan (u)

Substitute back

u = 4 x

= \frac{1}{4} arctan (4 x)

Add a constant to the solution

= \frac{1}{4} arctan (4 x) + C

\int_{- 1}^{2} (- y^{2} + y + 2) d y

x^{3} cos (x^{2})

\int \frac{x ^{2}}{25 - 9 x ^{2}} d x

2 y^{^{''}} - 9 y^{^{'}} - 5 y = 0

sec^{2} (π x)

What is the integral of 1/(1+16x^2) ?
The integral of 1/(1+16x^2) is 1/4 arctan(4x)+C