What is the integral of (3x)/(x^2+4x+5) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x)/(x^2+4x+5)

\int \frac{3 x}{x ^{2} + 4 x + 5} d x

3 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2)) + C

Solution steps

\int \frac{3 x}{x ^{2} + 4 x + 5} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{x ^{2} + 4 x + 5} d x

Complete the square

x^{2} + 4 x + 5 : (x + 2)^{2} + 1

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x + 2

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

Simplify

3 (\frac{1}{2} ln (x + 2)^{2} + 1 - 2 arctan (x + 2)) : 3 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2))

= 3 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2))

Add a constant to the solution

= 3 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2)) + C

\int 5 e^{x} sin (2 x) d x

\int 8 x (2 x - 1)^{4} d x

\int cos^{3} (x) + sin^{3} (x) d x

\int 5^{\frac{3}{4}} d x

\int (- x^{2} + x + 2) d x

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