What is the integral of (sqrt(A))/(MA-A^2) ?

The integral of (sqrt(A))/(MA-A^2) is 1/(sqrt(M))(ln|sqrt(A/M)+1|-ln|sqrt(A/M)-1|)+C

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integral of (sqrt(A))/(MA-A^2)

\int \frac{A}{M A - A ^{2}}

\frac{1}{M} (ln \frac{A}{M} + 1 - ln \frac{A}{M} - 1) + C

Solution steps

\int \frac{A}{M A - A ^{2}} d A

Apply u-substitution

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

= \int \frac{2}{- u ^{2} + M} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{- u ^{2} + M} d u

Apply Integral Substitution

= 2 \cdot \int \frac{1}{M ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{M} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 2 \cdot \frac{1}{M} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= 2 \cdot \frac{1}{M} \frac{ln \frac{A}{M} + 1}{2} - \frac{ln \frac{A}{M} - 1}{2}

Simplify

2 \cdot \frac{1}{M} \frac{ln \frac{A}{M} + 1}{2} - \frac{ln \frac{A}{M} - 1}{2} : \frac{1}{M} (ln \frac{A}{M} + 1 - ln \frac{A}{M} - 1)

= \frac{1}{M} (ln \frac{A}{M} + 1 - ln \frac{A}{M} - 1)

Add a constant to the solution

= \frac{1}{M} (ln \frac{A}{M} + 1 - ln \frac{A}{M} - 1) + C

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What is the integral of (sqrt(A))/(MA-A^2) ?
The integral of (sqrt(A))/(MA-A^2) is 1/(sqrt(M))(ln|sqrt(A/M)+1|-ln|sqrt(A/M)-1|)+C