What is the integral of (-5x)/(x^2+3x+4) ?

The integral of (-5x)/(x^2+3x+4) is -10(1/4 ln|4x^2+12x+16|-3/(2sqrt(7))arctan((2x+3)/(sqrt(7))))+C

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integral of (-5x)/(x^2+3x+4)

\int \frac{- 5 x}{x ^{2} + 3 x + 4} d x

- 10 (\frac{1}{4} ln 4 x^{2} + 12 x + 16 - \frac{3}{2 7} arctan (\frac{2 x + 3}{7})) + C

Solution steps

\int \frac{- 5 x}{x ^{2} + 3 x + 4} d x

Take the constant out:

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

= - 5 \cdot \int \frac{x}{x ^{2} + 3 x + 4} d x

Complete the square

x^{2} + 3 x + 4 : (x + \frac{3}{2})^{2} + \frac{7}{4}

= - 5 \cdot \int \frac{x}{( x + \frac{3}{2} ) ^{2} + \frac{7}{4}} d x

Apply u-substitution

= - 5 \cdot \int \frac{2 ( 2 u - 3 )}{4 u ^{2} + 7} d u

Take the constant out:

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

= - 5 \cdot 2 \cdot \int \frac{2 u - 3}{4 u ^{2} + 7} d u

Expand

\frac{2 u - 3}{4 u ^{2} + 7} : \frac{2 u}{4 u ^{2} + 7} - \frac{3}{4 u ^{2} + 7}

Apply the Sum Rule:

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

= - 5 \cdot 2 (\int \frac{2 u}{4 u ^{2} + 7} d u - \int \frac{3}{4 u ^{2} + 7} d u)

\int \frac{2 u}{4 u ^{2} + 7} d u = \frac{1}{4} ln 4 u^{2} + 7

\int \frac{3}{4 u ^{2} + 7} d u = \frac{3}{2 7} arctan (\frac{2}{7} u)

= - 5 \cdot 2 (\frac{1}{4} ln 4 u^{2} + 7 - \frac{3}{2 7} arctan (\frac{2}{7} u))

Substitute back

u = x + \frac{3}{2}

= - 5 \cdot 2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} + 7 - \frac{3}{2 7} arctan (\frac{2}{7} (x + \frac{3}{2})))

Simplify

- 5 \cdot 2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} + 7 - \frac{3}{2 7} arctan (\frac{2}{7} (x + \frac{3}{2}))) : - 10 (\frac{1}{4} ln 4 x^{2} + 12 x + 16 - \frac{3}{2 7} arctan (\frac{2 x + 3}{7}))

= - 10 (\frac{1}{4} ln 4 x^{2} + 12 x + 16 - \frac{3}{2 7} arctan (\frac{2 x + 3}{7}))

Add a constant to the solution

= - 10 (\frac{1}{4} ln 4 x^{2} + 12 x + 16 - \frac{3}{2 7} arctan (\frac{2 x + 3}{7})) + C

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What is the integral of (-5x)/(x^2+3x+4) ?
The integral of (-5x)/(x^2+3x+4) is -10(1/4 ln|4x^2+12x+16|-3/(2sqrt(7))arctan((2x+3)/(sqrt(7))))+C