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

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

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integral of 1/(xsqrt(a^2-a^2x^2))

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

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

Solution steps

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

a^{2} - a^{2} x^{2} = a^{2} 1 - x^{2}

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

Take the constant out:

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

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

Apply u-substitution

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

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

Substitute back

u = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{( y ^{3} )}{x ^{2}}

\frac{\partial}{\partial y} ((A^{2} x^{2} + B^{2} y^{2})^{4})

\int \frac{e ^{\frac{4}{x}}}{x ^{2}} d x

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