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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2} ln x^{2} + 6 x + 10 - arctan (x + 3) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

Substitute back

u = x + 3

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

Simplify

\frac{1}{2} ln (x + 3)^{2} + 1 - arctan (x + 3) : \frac{1}{2} ln x^{2} + 6 x + 10 - arctan (x + 3)

= \frac{1}{2} ln x^{2} + 6 x + 10 - arctan (x + 3)

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 6 x + 10 - arctan (x + 3) + C

x y^{^{'}} + 3 y = \frac{( e ^{x} )}{x}

cos^{3} (x) + sin^{3} (x)

\int sin^{2} (x co) s^{3} x d x

\frac{\partial}{\partial y} (5 x sin (y))

3 x^{2}

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