What is the integral of 1/(25a-a^3y^2) ?

The integral of 1/(25a-a^3y^2) is 1/(10a^2)(ln|(ay)/5+1|-ln|(ay)/5-1|)+C

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integral of 1/(25a-a^3y^2)

\int \frac{1}{25 a - a ^{3} y ^{2}} d y

\frac{1}{10 a ^{2}} (ln \frac{a y}{5} + 1 - ln \frac{a y}{5} - 1) + C

Solution steps

\int \frac{1}{25 a - a ^{3} y ^{2}} d y

Apply Integral Substitution

= \int \frac{1}{5 a ^{2} ( - 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}{5 a ^{2}} \cdot \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}{5 a ^{2}} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{a a}{5 a} y

= \frac{1}{5 a ^{2}} \frac{ln \frac{a a}{5 a} y + 1}{2} - \frac{ln \frac{a a}{5 a} y - 1}{2}

Simplify

\frac{1}{5 a ^{2}} \frac{ln \frac{a a}{5 a} y + 1}{2} - \frac{ln \frac{a a}{5 a} y - 1}{2} : \frac{1}{10 a ^{2}} (ln \frac{a y}{5} + 1 - ln \frac{a y}{5} - 1)

= \frac{1}{10 a ^{2}} (ln \frac{a y}{5} + 1 - ln \frac{a y}{5} - 1)

Add a constant to the solution

= \frac{1}{10 a ^{2}} (ln \frac{a y}{5} + 1 - ln \frac{a y}{5} - 1) + C

What is the integral of 1/(25a-a^3y^2) ?
The integral of 1/(25a-a^3y^2) is 1/(10a^2)(ln|(ay)/5+1|-ln|(ay)/5-1|)+C