What is the integral of 1/(3x^2-2x+4) ?

The integral of 1/(3x^2-2x+4) is 1/(sqrt(11))arctan((3x-1)/(sqrt(11)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(3x^2-2x+4)

\int \frac{1}{3 x ^{2} - 2 x + 4} d x

\frac{1}{11} arctan (\frac{3 x - 1}{11}) + C

Solution steps

\int \frac{1}{3 x ^{2} - 2 x + 4} d x

Complete the square

3 x^{2} - 2 x + 4 : 3 (x - \frac{1}{3})^{2} + \frac{11}{3}

= \int \frac{1}{3 ( x - \frac{1}{3} ) ^{2} + \frac{11}{3}} d x

Apply u-substitution

: \int \frac{3}{9 u ^{2} + 11} d u

= \int \frac{3}{9 u ^{2} + 11} d u

Take the constant out:

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

= 3 \cdot \int \frac{1}{9 u ^{2} + 11} d u

Apply Integral Substitution

= 3 \cdot \int \frac{1}{3 11 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 3 \cdot \frac{1}{3 11} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 3 \cdot \frac{1}{3 11} arctan (v)

Substitute back

= 3 \cdot \frac{1}{3 11} arctan (\frac{3}{11} (x - \frac{1}{3}))

Simplify

3 \cdot \frac{1}{3 11} arctan (\frac{3}{11} (x - \frac{1}{3})) : \frac{1}{11} arctan (\frac{3 x - 1}{11})

= \frac{1}{11} arctan (\frac{3 x - 1}{11})

Add a constant to the solution

= \frac{1}{11} arctan (\frac{3 x - 1}{11}) + C

y^{^{''}} + y = sec (t) tan (t)

2 - x^{2}, x, - 2, 1

n = 1 \sum \infty \frac{n !}{n ! + 1}

\int_{0}^{1} (\frac{x}{x ^{2} + 1}) d x

\int_{3}^{7} \frac{x}{x ^{2} + 6 x + 13} d x

What is the integral of 1/(3x^2-2x+4) ?
The integral of 1/(3x^2-2x+4) is 1/(sqrt(11))arctan((3x-1)/(sqrt(11)))+C