What is the integral of 5/(4x^2-4x+10) ?

The integral of 5/(4x^2-4x+10) is 5/6 arctan(2/3 x-1/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 5/(4x^2-4x+10)

\int \frac{5}{4 x ^{2} - 4 x + 10} d x

\frac{5}{6} arctan (\frac{2}{3} x - \frac{1}{3}) + C

Solution steps

\int \frac{5}{4 x ^{2} - 4 x + 10} 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}{4 x ^{2} - 4 x + 10} d x

Complete the square

4 x^{2} - 4 x + 10 : 4 (x - \frac{1}{2})^{2} + 9

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

Apply u-substitution

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

Apply Integral Substitution

= 5 \cdot \int \frac{1}{6 ( 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}{6} \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}{6} arctan (v)

Substitute back

= 5 \cdot \frac{1}{6} arctan (\frac{2}{3} (x - \frac{1}{2}))

Simplify

5 \cdot \frac{1}{6} arctan (\frac{2}{3} (x - \frac{1}{2})) : \frac{5}{6} arctan (\frac{2}{3} x - \frac{1}{3})

= \frac{5}{6} arctan (\frac{2}{3} x - \frac{1}{3})

Add a constant to the solution

= \frac{5}{6} arctan (\frac{2}{3} x - \frac{1}{3}) + C

What is the integral of 5/(4x^2-4x+10) ?
The integral of 5/(4x^2-4x+10) is 5/6 arctan(2/3 x-1/3)+C