What is the integral of (2x+1)/(x^2+6x+10) ?

The integral of (2x+1)/(x^2+6x+10) is ln|x^2+6x+10|-5arctan(x+3)+C

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integral of (2x+1)/(x^2+6x+10)

\int \frac{2 x + 1}{x ^{2} + 6 x + 10} d x

ln x^{2} + 6 x + 10 - 5 arctan (x + 3) + C

Solution steps

\int \frac{2 x + 1}{x ^{2} + 6 x + 10} d x

Complete the square

x^{2} + 6 x + 10 : (x + 3)^{2} + 1

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

Apply u-substitution

: \int \frac{2 u - 5}{u ^{2} + 1} d u

= \int \frac{2 u - 5}{u ^{2} + 1} d u

Expand

\frac{2 u - 5}{u ^{2} + 1} : \frac{2 u}{u ^{2} + 1} - \frac{5}{u ^{2} + 1}

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} + 1} d u - \int \frac{5}{u ^{2} + 1} d u

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

\int \frac{5}{u ^{2} + 1} d u = 5 arctan (u)

= ln u^{2} + 1 - 5 arctan (u)

Substitute back

u = x + 3

= ln (x + 3)^{2} + 1 - 5 arctan (x + 3)

Expand

(x + 3)^{2} + 1 : x^{2} + 6 x + 10

= ln x^{2} + 6 x + 10 - 5 arctan (x + 3)

Add a constant to the solution

= ln x^{2} + 6 x + 10 - 5 arctan (x + 3) + C

t \to \infty lim (\frac{1}{4 t})

f (x) = \frac{x}{x + 6}

ln (x y)

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

\int (x + 1)^{4} d x

What is the integral of (2x+1)/(x^2+6x+10) ?
The integral of (2x+1)/(x^2+6x+10) is ln|x^2+6x+10|-5arctan(x+3)+C