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

The integral of 1/(a-x^2) is 1/(2sqrt(a))(ln| x/(sqrt(a))+1|-ln| x/(sqrt(a))-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/(a-x^2)

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

\frac{1}{2 a} (ln \frac{x}{a} + 1 - ln \frac{x}{a} - 1) + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = \frac{x}{a}

= \frac{1}{a} \frac{ln \frac{x}{a} + 1}{2} - \frac{ln \frac{x}{a} - 1}{2}

Simplify

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

= \frac{1}{2 a} (ln \frac{x}{a} + 1 - ln \frac{x}{a} - 1)

Add a constant to the solution

= \frac{1}{2 a} (ln \frac{x}{a} + 1 - ln \frac{x}{a} - 1) + C

x^{3} + 12 x^{2} + 54 x + 14

\int sec^{2} (x) \cdot tan^{5} (x) d x

\frac{d ^{2}}{d x ^{2}} ((5 x^{2} + 4 x) cos (x))

\int_{4}^{5} x^{3} d x

\int t^{15} e^{- t^{8}} d t

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