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

The integral of x/(2x^2+2x+7) is 1/4 ln|4x^2+4x+14|-1/(2sqrt(13))arctan((2x+1)/(sqrt(13)))+C

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

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

\frac{1}{4} ln 4 x^{2} + 4 x + 14 - \frac{1}{2 13} arctan (\frac{2 x + 1}{13}) + C

Solution steps

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

Complete the square

2 x^{2} + 2 x + 7 : 2 (x + \frac{1}{2})^{2} + \frac{13}{2}

= \int \frac{x}{2 ( x + \frac{1}{2} ) ^{2} + \frac{13}{2}} d x

Apply u-substitution

: \int \frac{2 u - 1}{4 u ^{2} + 13} d u

= \int \frac{2 u - 1}{4 u ^{2} + 13} d u

Expand

\frac{2 u - 1}{4 u ^{2} + 13} : \frac{2 u}{4 u ^{2} + 13} - \frac{1}{4 u ^{2} + 13}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 u}{4 u ^{2} + 13} d u - \int \frac{1}{4 u ^{2} + 13} d u

\int \frac{2 u}{4 u ^{2} + 13} d u = \frac{1}{4} ln 4 u^{2} + 13

\int \frac{1}{4 u ^{2} + 13} d u = \frac{1}{2 13} arctan (\frac{2}{13} u)

= \frac{1}{4} ln 4 u^{2} + 13 - \frac{1}{2 13} arctan (\frac{2}{13} u)

Substitute back

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

= \frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 13 - \frac{1}{2 13} arctan (\frac{2}{13} (x + \frac{1}{2}))

Simplify

\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 13 - \frac{1}{2 13} arctan (\frac{2}{13} (x + \frac{1}{2})) : \frac{1}{4} ln 4 x^{2} + 4 x + 14 - \frac{1}{2 13} arctan (\frac{2 x + 1}{13})

= \frac{1}{4} ln 4 x^{2} + 4 x + 14 - \frac{1}{2 13} arctan (\frac{2 x + 1}{13})

Add a constant to the solution

= \frac{1}{4} ln 4 x^{2} + 4 x + 14 - \frac{1}{2 13} arctan (\frac{2 x + 1}{13}) + C

y^{^{''}} + a^{2} y = sin (b \cdot t)

\int (6 x^{2} - 4 x + 2) (x^{3} - x^{2} + x + 1)^{7} d x

\frac{\partial}{\partial x} (a^{2} - x^{2} - y^{2})

\int \frac{e ^{x^{2}}}{x} d x

x \to 3 lim (\frac{5}{2 x - 6})

What is the integral of x/(2x^2+2x+7) ?
The integral of x/(2x^2+2x+7) is 1/4 ln|4x^2+4x+14|-1/(2sqrt(13))arctan((2x+1)/(sqrt(13)))+C