What is the integral of (x+1)/(x^2+3x+4) ?

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

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integral of (x+1)/(x^2+3x+4)

\int \frac{x + 1}{x ^{2} + 3 x + 4} d x

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

Solution steps

\int \frac{x + 1}{x ^{2} + 3 x + 4} d x

Complete the square

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

= \int \frac{x + 1}{( x + \frac{3}{2} ) ^{2} + \frac{7}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u - 1 )}{4 u ^{2} + 7} d u

= \int \frac{2 ( 2 u - 1 )}{4 u ^{2} + 7} d u

Take the constant out:

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

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

Expand

\frac{2 u - 1}{4 u ^{2} + 7} : \frac{2 u}{4 u ^{2} + 7} - \frac{1}{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

= 2 (\int \frac{2 u}{4 u ^{2} + 7} d u - \int \frac{1}{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{1}{4 u ^{2} + 7} d u = \frac{1}{2 7} arctan (\frac{2}{7} u)

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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