What is the integral of 5/(4+16x^2) ?

The integral of 5/(4+16x^2) is 5/8 arctan(2x)+C

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integral of 5/(4+16x^2)

\int \frac{5}{4 + 16 x ^{2}} d x

\frac{5}{8} arctan (2 x) + C

Solution steps

\int \frac{5}{4 + 16 x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{4 + 16 x ^{2}} d x

Apply Integral Substitution

= 5 \cdot \int \frac{1}{2 ( 4 + 4 u ^{2} )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{4 + 4 u ^{2}} d u

\frac{1}{4 + 4 u ^{2}} = \frac{1}{4} \frac{1}{1 + u ^{2}}

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{4} \cdot \frac{1}{1 + u ^{2}} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \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)

= 5 \cdot \frac{1}{2} \cdot \frac{1}{4} arctan (u)

Substitute back

u = \frac{4}{2} x

= 5 \cdot \frac{1}{2} \cdot \frac{1}{4} arctan (\frac{4}{2} x)

Simplify

5 \cdot \frac{1}{2} \cdot \frac{1}{4} arctan (\frac{4}{2} x) : \frac{5}{8} arctan (2 x)

= \frac{5}{8} arctan (2 x)

Add a constant to the solution

= \frac{5}{8} arctan (2 x) + C

What is the integral of 5/(4+16x^2) ?
The integral of 5/(4+16x^2) is 5/8 arctan(2x)+C