What is the integral of 1/(2x^2-2x+1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(2x^2-2x+1)

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

arctan (2 x - 1) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 2 \cdot \frac{1}{2} arctan (v)

Substitute back

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

Simplify

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

= arctan (2 x - 1)

Add a constant to the solution

= arctan (2 x - 1) + C

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

f (x) = \frac{1}{x - 2}, at x = 7

\int \frac{x}{36 x ^{2} - 1} d x

f (x) = 6 e^{x} + \frac{2}{3 x}

\frac{\partial}{\partial x} (ln (x^{7} + y^{2}))

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