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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

\int \frac{5}{4 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

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

Complete the square

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

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

Apply u-substitution

: \int \frac{1}{4 u ^{2} + 4} d u

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

\frac{1}{4 u ^{2} + 4} = \frac{1}{4} \frac{1}{u ^{2} + 1}

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

Take the constant out:

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

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

Use the common integral:

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

= 5 \cdot \frac{1}{4} arctan (u)

Substitute back

u = x - \frac{1}{2}

= 5 \cdot \frac{1}{4} arctan (x - \frac{1}{2})

Simplify

5 \cdot \frac{1}{4} arctan (x - \frac{1}{2}) : \frac{5}{4} arctan (x - \frac{1}{2})

= \frac{5}{4} arctan (x - \frac{1}{2})

Add a constant to the solution

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

\int [\frac{6 x ^{2} + x - 3}{x ^{3} - x}] d x

\int \frac{2 - x ^{2}}{6 x - x ^{3}} d x

\int \frac{3 \cdot x - 4}{x ^{2} - x - 2} d x

\int cos^{2} (39 x) d x

\int 4^{5 t} d t

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