What is the integral of 1/(y^2-7) ?

The integral of 1/(y^2-7) is -1/(2sqrt(7))(ln| y/(sqrt(7))+1|-ln| y/(sqrt(7))-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/(y^2-7)

\int \frac{1}{y ^{2} - 7} d y

- \frac{1}{2 7} (ln \frac{y}{7} + 1 - ln \frac{y}{7} - 1) + C

Solution steps

\int \frac{1}{y ^{2} - 7} d y

Apply Integral Substitution

= \int \frac{1}{7 ( u ^{2} - 1 )} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \frac{1}{7} \cdot \int \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

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

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{7} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{y}{7}

= \frac{1}{7} - \frac{ln \frac{y}{7} + 1}{2} - \frac{ln \frac{y}{7} - 1}{2}

Simplify

\frac{1}{7} - \frac{ln \frac{y}{7} + 1}{2} - \frac{ln \frac{y}{7} - 1}{2} : - \frac{1}{2 7} (ln \frac{y}{7} + 1 - ln \frac{y}{7} - 1)

= - \frac{1}{2 7} (ln \frac{y}{7} + 1 - ln \frac{y}{7} - 1)

Add a constant to the solution

= - \frac{1}{2 7} (ln \frac{y}{7} + 1 - ln \frac{y}{7} - 1) + C

\int \frac{1}{r ^{2} 4 r ^{2} - 25} d r

\int \frac{1}{sec ( x )} d x

\int sec (2 w) tan (2 w) d w

\int (4 y + 2 x - 5) d x

\int (x^{5} + 1) cos (x^{6} + 6 x) d x

What is the integral of 1/(y^2-7) ?
The integral of 1/(y^2-7) is -1/(2sqrt(7))(ln| y/(sqrt(7))+1|-ln| y/(sqrt(7))-1|)+C