What is the integral of (2x+5)/(x^2+4x+7) ?

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

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integral of (2x+5)/(x^2+4x+7)

\int \frac{2 x + 5}{x ^{2} + 4 x + 7} d x

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

Solution steps

\int \frac{2 x + 5}{x ^{2} + 4 x + 7} d x

Complete the square

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

= \int \frac{2 x + 5}{( x + 2 ) ^{2} + 3} d x

Apply u-substitution

: \int \frac{2 u + 1}{u ^{2} + 3} d u

= \int \frac{2 u + 1}{u ^{2} + 3} d u

Expand

\frac{2 u + 1}{u ^{2} + 3} : \frac{2 u}{u ^{2} + 3} + \frac{1}{u ^{2} + 3}

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}{u ^{2} + 3} d u + \int \frac{1}{u ^{2} + 3} d u

\int \frac{2 u}{u ^{2} + 3} d u = ln u^{2} + 3

\int \frac{1}{u ^{2} + 3} d u = \frac{1}{3} arctan (\frac{u}{3})

= ln u^{2} + 3 + \frac{1}{3} arctan (\frac{u}{3})

Substitute back

u = x + 2

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

Simplify

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

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

Add a constant to the solution

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

5 x^{3}

\int - 2 x e^{- 4 x} d x

\int \frac{1}{3 ( x - 1 ) ^{4}} d x

\int \frac{5}{t + 4} d t

\int 5 t^{5} d t

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