What is the integral of 8/(9+r^2) ?

The integral of 8/(9+r^2) is 8/3 arctan(r/3)+C

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integral of 8/(9+r^2)

\int \frac{8}{9 + r ^{2}} d r

\frac{8}{3} arctan (\frac{r}{3}) + C

Solution steps

\int \frac{8}{9 + r ^{2}} d r

Take the constant out:

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

= 8 \cdot \int \frac{1}{9 + r ^{2}} d r

Apply Integral Substitution

= 8 \cdot \int \frac{1}{3 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 8 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 8 \cdot \frac{1}{3} arctan (u)

Substitute back

u = \frac{r}{3}

= 8 \cdot \frac{1}{3} arctan (\frac{r}{3})

Simplify

8 \cdot \frac{1}{3} arctan (\frac{r}{3}) : \frac{8}{3} arctan (\frac{r}{3})

= \frac{8}{3} arctan (\frac{r}{3})

Add a constant to the solution

= \frac{8}{3} arctan (\frac{r}{3}) + C

f (x) = \frac{( 4 x ) ^{3}}{( 3 - 5 x ) ^{5}}

\int 4 x ln (4 x) d x

ln (e^{x} x)

\frac{1}{1 - x}, (\frac{π}{4})

\int \frac{( 2 x + 1 )}{( x + 2 ) ^{4} ( x - 1 ) ^{4}} d x

What is the integral of 8/(9+r^2) ?
The integral of 8/(9+r^2) is 8/3 arctan(r/3)+C