What is the integral of ((x+4))/(x^2+8x+17) ?

The integral of ((x+4))/(x^2+8x+17) is 1/2 ln|x^2+8x+17|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x+4))/(x^2+8x+17)

\int \frac{( x + 4 )}{x ^{2} + 8 x + 17} d x

\frac{1}{2} ln x^{2} + 8 x + 17 + C

Solution steps

\int \frac{x + 4}{x ^{2} + 8 x + 17} d x

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{2 v} d v

= \int \frac{1}{2 v} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

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

Simplify

\frac{1}{2} ln (x + 4)^{2} + 1 : \frac{1}{2} ln x^{2} + 8 x + 17

= \frac{1}{2} ln x^{2} + 8 x + 17

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 8 x + 17 + C

y^{^{'}} + \frac{2}{t} y = \frac{cos ( t )}{t ^{2}}, y (π) = 0

\frac{1}{3 x + 5}

\int - 25 cos (- 5 t) d t

x \to - 1 lim (\frac{2}{x + 1})

What is the integral of ((x+4))/(x^2+8x+17) ?
The integral of ((x+4))/(x^2+8x+17) is 1/2 ln|x^2+8x+17|+C