What is the integral of 5/(x^2-4x+9) ?

The integral of 5/(x^2-4x+9) is sqrt(5)arctan(1/(sqrt(5))(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 5/(x^2-4x+9)

\int \frac{5}{x ^{2} - 4 x + 9} d x

5 arctan (\frac{1}{5} (x - 2)) + C

Solution steps

\int \frac{5}{x ^{2} - 4 x + 9} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ^{2} - 4 x + 9} d x

Complete the square

x^{2} - 4 x + 9 : (x - 2)^{2} + 5

= 5 \cdot \int \frac{1}{( x - 2 ) ^{2} + 5} d x

Apply u-substitution

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

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

Apply Integral Substitution

= 5 \cdot \int \frac{1}{5 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 5 \cdot \frac{1}{5} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 5 \cdot \frac{1}{5} arctan (v)

Substitute back

= 5 \cdot \frac{1}{5} arctan (\frac{x - 2}{5})

Simplify

5 \cdot \frac{1}{5} arctan (\frac{x - 2}{5}) : 5 arctan (\frac{1}{5} (x - 2))

= 5 arctan (\frac{1}{5} (x - 2))

Add a constant to the solution

= 5 arctan (\frac{1}{5} (x - 2)) + C

\int \frac{x ^{5}}{( x ^{3} + 1 ) ^{5}} d x

\frac{1}{cos ( x )}

\int x^{2} x^{3} - 15 d x

\int sin (5 x) e^{c o s (5 x)} d x

\int \frac{1}{7 x ^{2} + 2} d x

What is the integral of 5/(x^2-4x+9) ?
The integral of 5/(x^2-4x+9) is sqrt(5)arctan(1/(sqrt(5))(x-2))+C