What is the integral of (130)/((x+4)^2) ?

The integral of (130)/((x+4)^2) is -(130)/(x+4)+C

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integral of (130)/((x+4)^2)

\int \frac{130}{( x + 4 ) ^{2}} d x

- \frac{130}{x + 4} + C

Solution steps

\int \frac{130}{( x + 4 ) ^{2}} d x

Take the constant out:

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

= 130 \cdot \int \frac{1}{( x + 4 ) ^{2}} d x

Apply u-substitution

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

= 130 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 130 (- \frac{1}{u})

Substitute back

u = x + 4

= 130 (- \frac{1}{x + 4})

Simplify

130 (- \frac{1}{x + 4}) : - \frac{130}{x + 4}

= - \frac{130}{x + 4}

Add a constant to the solution

= - \frac{130}{x + 4} + C

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What is the integral of (130)/((x+4)^2) ?
The integral of (130)/((x+4)^2) is -(130)/(x+4)+C