What is the integral of 1/(25-36x^2) ?

The integral of 1/(25-36x^2) is 1/60 (ln| 6/5 x+1|-ln| 6/5 x-1|)+C

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integral of 1/(25-36x^2)

\int \frac{1}{25 - 36 x ^{2}} d x

\frac{1}{60} (ln \frac{6}{5} x + 1 - ln \frac{6}{5} x - 1) + C

Solution steps

\int \frac{1}{25 - 36 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{30 ( - 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}{30} \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}{30} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{6}{5} x

= \frac{1}{30} (\frac{ln \frac{6}{5} x + 1}{2} - \frac{ln \frac{6}{5} x - 1}{2})

Simplify

\frac{1}{30} (\frac{ln \frac{6}{5} x + 1}{2} - \frac{ln \frac{6}{5} x - 1}{2}) : \frac{1}{60} (ln \frac{6}{5} x + 1 - ln \frac{6}{5} x - 1)

= \frac{1}{60} (ln \frac{6}{5} x + 1 - ln \frac{6}{5} x - 1)

Add a constant to the solution

= \frac{1}{60} (ln \frac{6}{5} x + 1 - ln \frac{6}{5} x - 1) + C

n = 4 \sum \infty e^{5 - 6 n}

\frac{\partial}{\partial y} (y (1 + ln (x y)))

t y^{^{'}} + t y = 1 - y, y (1) = 0

\int \frac{1}{x - 1} d x

(x^{3} - 5 x) (4 x^{2} - 3 x)

What is the integral of 1/(25-36x^2) ?
The integral of 1/(25-36x^2) is 1/60 (ln| 6/5 x+1|-ln| 6/5 x-1|)+C