What is the integral of 3/(1+4x^2) ?

The integral of 3/(1+4x^2) is 3/2 arctan(2x)+C

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integral of 3/(1+4x^2)

\int \frac{3}{1 + 4 x ^{2}} d x

\frac{3}{2} arctan (2 x) + C

Solution steps

\int \frac{3}{1 + 4 x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{1 + 4 x ^{2}} d x

Apply Integral Substitution

= 3 \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

= 3 \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)

= 3 \cdot \frac{1}{2} arctan (u)

Substitute back

u = 2 x

= 3 \cdot \frac{1}{2} arctan (2 x)

Simplify

3 \cdot \frac{1}{2} arctan (2 x) : \frac{3}{2} arctan (2 x)

= \frac{3}{2} arctan (2 x)

Add a constant to the solution

= \frac{3}{2} arctan (2 x) + C

\int 3 x^{2} (2 x^{2} - 5) d x

\int (\frac{6}{x}) d x

\int 2 x + 6 e^{x} d x

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

y^{^{''}} - 2 y^{^{'}} + y = e^{x} ln (x)

What is the integral of 3/(1+4x^2) ?
The integral of 3/(1+4x^2) is 3/2 arctan(2x)+C