What is the integral of 1/(2x^2+2x+5) ?

The integral of 1/(2x^2+2x+5) is 1/3 arctan(2/3 x+1/3)+C

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integral of 1/(2x^2+2x+5)

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

\frac{1}{3} arctan (\frac{2}{3} x + \frac{1}{3}) + C

Solution steps

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

Complete the square

2 x^{2} + 2 x + 5 : 2 (x + \frac{1}{2})^{2} + \frac{9}{2}

= \int \frac{1}{2 ( x + \frac{1}{2} ) ^{2} + \frac{9}{2}} d x

Apply u-substitution

: \int \frac{2}{4 u ^{2} + 9} d u

= \int \frac{2}{4 u ^{2} + 9} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{4 u ^{2} + 9} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{6 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 2 \cdot \frac{1}{6} arctan (v)

Substitute back

= 2 \cdot \frac{1}{6} arctan (\frac{2}{3} (x + \frac{1}{2}))

Simplify

2 \cdot \frac{1}{6} arctan (\frac{2}{3} (x + \frac{1}{2})) : \frac{1}{3} arctan (\frac{2}{3} x + \frac{1}{3})

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x^{\frac{3}{4}} \cdot y^{- 4})

\frac{d}{d x} (4 arcsin (x^{4}))

n = 0 \sum \infty \frac{6 ^{n}}{7 ^{n}}

x \to 0 + lim (ln (x) sin (x))

y^{^{''}} + 7 y^{^{'}} = 0, y (0) = - 4, y^{^{'}} (0) = 7

What is the integral of 1/(2x^2+2x+5) ?
The integral of 1/(2x^2+2x+5) is 1/3 arctan(2/3 x+1/3)+C