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

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

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

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

arctan (2 x) + C

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 2 x

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

Simplify

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

= arctan (2 x)

Add a constant to the solution

= arctan (2 x) + C

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

(1, \frac{13}{2}), (6, - 4)

\frac{d}{d x} (\frac{4 x + 6}{x ^{2} + 3 x + 4})

\frac{2}{( s ^{2} + 1 ) ^{2}}

y = \frac{x + \frac{1}{x}}{2 x + \frac{1}{x}}