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

The integral of (16)/(1+4x^2) is 8arctan(2x)+C

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

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

8 arctan (2 x) + C

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= 16 \cdot \int \frac{1}{2 ( 1 + u ^{2} )} d u

Take the constant out:

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

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

Use the common integral:

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

= 16 \cdot \frac{1}{2} arctan (u)

Substitute back

u = 2 x

= 16 \cdot \frac{1}{2} arctan (2 x)

Simplify

16 \cdot \frac{1}{2} arctan (2 x) : 8 arctan (2 x)

= 8 arctan (2 x)

Add a constant to the solution

= 8 arctan (2 x) + C

What is the integral of (16)/(1+4x^2) ?
The integral of (16)/(1+4x^2) is 8arctan(2x)+C