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

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

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integral of (r^2)/((a^2-r^2))

integral

\frac{r ^{2}}{( a ^{2} - r ^{2} )}

\frac{a}{2} ln \frac{r}{a} + 1 - \frac{a}{2} ln \frac{r}{a} - 1 - r + C

Solution steps

\int \frac{r ^{2}}{a ^{2} - r ^{2}} d r

Apply Integral Substitution

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

Take the constant out:

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

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

\frac{u ^{2}}{1 - u ^{2}} = \frac{1}{1 - u ^{2}} - 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= a (\int \frac{1}{1 - u ^{2}} d u - \int 1 d u)

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

\int 1 d u = u

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

Substitute back

u = \frac{r}{a}

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

Simplify

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

= \frac{a}{2} ln \frac{r}{a} + 1 - \frac{a}{2} ln \frac{r}{a} - 1 - r

Add a constant to the solution

= \frac{a}{2} ln \frac{r}{a} + 1 - \frac{a}{2} ln \frac{r}{a} - 1 - r + C

\frac{2}{( 1 - x ^{2} )}

\frac{2}{( x \cdot 1 + x )}

\frac{( 6 x - 4 )}{( x ^{3} - 4 x )}

\frac{( x ^{2} + x )}{( x ^{2} + 1 ) ^{2}}

\frac{c}{r ^{3}}

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