What is the integral of 1/(7x^2-14x+49) ?

The integral of 1/(7x^2-14x+49) is 1/(7sqrt(6))arctan(1/(sqrt(6))(x-1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(7x^2-14x+49)

\int \frac{1}{7 x ^{2} - 14 x + 49} d x

\frac{1}{7 6} arctan (\frac{1}{6} (x - 1)) + C

Solution steps

\int \frac{1}{7 x ^{2} - 14 x + 49} d x

Complete the square

7 x^{2} - 14 x + 49 : 7 (x - 1)^{2} + 42

= \int \frac{1}{7 ( x - 1 ) ^{2} + 42} d x

Apply u-substitution

= \int \frac{1}{7 u ^{2} + 42} d u

Apply Integral Substitution

= \int \frac{1}{7 6 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{7 6} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{7 6} arctan (v)

Substitute back

= \frac{1}{7 6} arctan (\frac{7}{42} (x - 1))

Simplify

\frac{1}{7 6} arctan (\frac{7}{42} (x - 1)) : \frac{1}{7 6} arctan (\frac{1}{6} (x - 1))

= \frac{1}{7 6} arctan (\frac{1}{6} (x - 1))

Add a constant to the solution

= \frac{1}{7 6} arctan (\frac{1}{6} (x - 1)) + C

What is the integral of 1/(7x^2-14x+49) ?
The integral of 1/(7x^2-14x+49) is 1/(7sqrt(6))arctan(1/(sqrt(6))(x-1))+C