What is the integral of 5/((2x-3)^2+9) ?

The integral of 5/((2x-3)^2+9) is 5/6 arctan(1/3 (2x-3))+C

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integral of 5/((2x-3)^2+9)

\int \frac{5}{( 2 x - 3 ) ^{2} + 9} d x

\frac{5}{6} arctan (\frac{1}{3} (2 x - 3)) + C

Solution steps

\int \frac{5}{( 2 x - 3 ) ^{2} + 9} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{( 2 x - 3 ) ^{2} + 9} d x

Apply u-substitution

= 5 \cdot \int \frac{1}{2 ( u ^{2} + 9 )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

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

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

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

Substitute back

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

Simplify

5 \cdot \frac{1}{2} \cdot \frac{1}{3} arctan (\frac{2 x - 3}{3}) : \frac{5}{6} arctan (\frac{1}{3} (2 x - 3))

= \frac{5}{6} arctan (\frac{1}{3} (2 x - 3))

Add a constant to the solution

= \frac{5}{6} arctan (\frac{1}{3} (2 x - 3)) + C

What is the integral of 5/((2x-3)^2+9) ?
The integral of 5/((2x-3)^2+9) is 5/6 arctan(1/3 (2x-3))+C