What is the integral of (32)/(1+16x^2) ?

The integral of (32)/(1+16x^2) is 8arctan(4x)+C

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integral of (32)/(1+16x^2)

\int \frac{32}{1 + 16 x ^{2}} d x

8 arctan (4 x) + C

Solution steps

\int \frac{32}{1 + 16 x ^{2}} d x

Take the constant out:

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

= 32 \cdot \int \frac{1}{1 + 16 x ^{2}} d x

Apply Integral Substitution

= 32 \cdot \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

= 32 \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)

= 32 \cdot \frac{1}{4} arctan (u)

Substitute back

u = 4 x

= 32 \cdot \frac{1}{4} arctan (4 x)

Simplify

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

= 8 arctan (4 x)

Add a constant to the solution

= 8 arctan (4 x) + C

n = 1 \sum \infty (\frac{1}{5})^{n - 1}

t + 4

10 e^{- 10 t}

f (x) = ln (x^{4})

\int cos (\frac{1}{3} x) d x

What is the integral of (32)/(1+16x^2) ?
The integral of (32)/(1+16x^2) is 8arctan(4x)+C