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

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

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

\int \frac{1}{( 2 x + 1 ) ^{2} + 2} d x

\frac{1}{2 2} arctan (\frac{1}{2} (2 x + 1)) + C

Solution steps

\int \frac{1}{( 2 x + 1 ) ^{2} + 2} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2 2} arctan (\frac{1}{2} (2 x + 1)) + C

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