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

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

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integral of 8/(9+x^2)

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

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

Solution steps

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

Take the constant out:

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

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

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{x}{3}

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + 4 y^{^{'}} - 5 y = 0

(x + 2)^{2} \frac{d y}{d x} = 5 - 8 y - 4 x y

x \to 3 - lim (\frac{- 5}{x - 3})

\int (2 + cos (x))^{2} d x

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

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