What is the integral of 1/(2x^2+4x+7) ?

The integral of 1/(2x^2+4x+7) is 1/(sqrt(10))arctan(sqrt(2/5)(x+1))+C

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integral of 1/(2x^2+4x+7)

\int \frac{1}{2 x ^{2} + 4 x + 7} d x

\frac{1}{10} arctan (\frac{2}{5} (x + 1)) + C

Solution steps

\int \frac{1}{2 x ^{2} + 4 x + 7} d x

Complete the square

2 x^{2} + 4 x + 7 : 2 (x + 1)^{2} + 5

= \int \frac{1}{2 ( x + 1 ) ^{2} + 5} d x

Apply u-substitution

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

Apply Integral Substitution

= \int \frac{1}{10 ( 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}{10} \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}{10} arctan (v)

Substitute back

= \frac{1}{10} arctan (\frac{2}{5} (x + 1))

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{1}{10} arctan (\frac{2}{5} (x + 1))

Add a constant to the solution

= \frac{1}{10} arctan (\frac{2}{5} (x + 1)) + C

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What is the integral of 1/(2x^2+4x+7) ?
The integral of 1/(2x^2+4x+7) is 1/(sqrt(10))arctan(sqrt(2/5)(x+1))+C