What is the integral of 3/((x+5)^2+4) ?

The integral of 3/((x+5)^2+4) is 3/2 arctan(1/2 (x+5))+C

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integral of 3/((x+5)^2+4)

\int \frac{3}{( x + 5 ) ^{2} + 4} d x

\frac{3}{2} arctan (\frac{1}{2} (x + 5)) + C

Solution steps

\int \frac{3}{( x + 5 ) ^{2} + 4} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{( x + 5 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 4} d u

= 3 \cdot \int \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 3 \cdot \frac{1}{2} arctan (v)

Substitute back

= 3 \cdot \frac{1}{2} arctan (\frac{x + 5}{2})

Simplify

3 \cdot \frac{1}{2} arctan (\frac{x + 5}{2}) : \frac{3}{2} arctan (\frac{1}{2} (x + 5))

= \frac{3}{2} arctan (\frac{1}{2} (x + 5))

Add a constant to the solution

= \frac{3}{2} arctan (\frac{1}{2} (x + 5)) + C

\frac{d}{d x} (e sin (x))

(\frac{1}{y ^{2}} - \frac{3}{y ^{4}}) (y + 9 y^{3})

h \to 0 lim (\frac{( x + h ) ^{2} - 9}{h})

x^{3}, - x, 1

n = 0 \sum \infty cos (\frac{1}{n})

What is the integral of 3/((x+5)^2+4) ?
The integral of 3/((x+5)^2+4) is 3/2 arctan(1/2 (x+5))+C