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

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

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

- \frac{1}{2} ln 1 - y^{2} + 1 + \frac{1}{2} ln 1 - y^{2} - 1 + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{- u ^{2} + 1} d u

= \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

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

Substitute back

u = 1 - y^{2}

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

Simplify

- \frac{ln 1 - y ^{2} + 1}{2} - \frac{ln 1 - y ^{2} - 1}{2} : - \frac{1}{2} ln 1 - y^{2} + 1 + \frac{1}{2} ln 1 - y^{2} - 1

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

Add a constant to the solution

= - \frac{1}{2} ln 1 - y^{2} + 1 + \frac{1}{2} ln 1 - y^{2} - 1 + C

\int 3 4 - x^{2} d x

\int sec^{2} (6 x) tan^{2} (6 x) d x

\int 8 t e^{2 t} d t

\int \frac{1 + ln ( x )}{3 + x ln ( x )} d x

\int \frac{3 x + 5}{2 x + 3} d x

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