What is the integral of 1/(4x^2-6x+10) ?

The integral of 1/(4x^2-6x+10) is 1/(sqrt(31))arctan((4x-3)/(sqrt(31)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(4x^2-6x+10)

\int \frac{1}{4 x ^{2} - 6 x + 10} d x

\frac{1}{31} arctan (\frac{4 x - 3}{31}) + C

Solution steps

\int \frac{1}{4 x ^{2} - 6 x + 10} d x

Complete the square

4 x^{2} - 6 x + 10 : 4 (x - \frac{3}{4})^{2} + \frac{31}{4}

= \int \frac{1}{4 ( x - \frac{3}{4} ) ^{2} + \frac{31}{4}} d x

Apply u-substitution

: \int \frac{4}{16 u ^{2} + 31} d u

= \int \frac{4}{16 u ^{2} + 31} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{16 u ^{2} + 31} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{4 31 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{4 31} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 4 \cdot \frac{1}{4 31} arctan (v)

Substitute back

= 4 \cdot \frac{1}{4 31} arctan (\frac{4}{31} (x - \frac{3}{4}))

Simplify

4 \cdot \frac{1}{4 31} arctan (\frac{4}{31} (x - \frac{3}{4})) : \frac{1}{31} arctan (\frac{4 x - 3}{31})

= \frac{1}{31} arctan (\frac{4 x - 3}{31})

Add a constant to the solution

= \frac{1}{31} arctan (\frac{4 x - 3}{31}) + C

\int 2 x cos (7 x) d x

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

\int sin (uv) d u

\int tan^{2} (x) tan^{2} (x) d x

\int (2 x - x^{\frac{1}{2}})^{3} d x

What is the integral of 1/(4x^2-6x+10) ?
The integral of 1/(4x^2-6x+10) is 1/(sqrt(31))arctan((4x-3)/(sqrt(31)))+C