What is the integral of [ 4/(16x^2+1)] ?

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

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integral of [ 4/(16x^2+1)]

\int [\frac{4}{16 x ^{2} + 1}] d x

arctan (4 x) + C

Solution steps

\int \frac{4}{16 x ^{2} + 1} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{16 x ^{2} + 1} d x

Apply Integral Substitution

= 4 \cdot \int \frac{1}{4 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

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

Substitute back

u = 4 x

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

Simplify

4 \cdot \frac{1}{4} arctan (4 x) : arctan (4 x)

= arctan (4 x)

Add a constant to the solution

= arctan (4 x) + C

f (x) = x^{2} - 5 x + 5, (6, 11)

\frac{\partial}{\partial s} (s + t)

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

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

\frac{d}{d x} (ln (x^{3}))

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