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

The integral of 1/(a-bx^2) is 1/(2sqrt(a)\sqrt{b)}(ln|sqrt(b/a)x+1|-ln|sqrt(b/a)x-1|)+C

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integral of 1/(a-bx^2)

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

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

Solution steps

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

Apply Integral Substitution

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

Substitute back

u = \frac{b}{a} x

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

Simplify

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

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

Add a constant to the solution

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

\int 1 + 5 x^{2} d x

\int \frac{3 x + 2}{( x - 4 ) ( x + 2 )} d x

e^{t}

\int (5 x + 1) (2 x - 2) d x

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