What is the integral of (15)/((0.5t+1)^2) ?

The integral of (15)/((0.5t+1)^2) is -(30)/(0.5t+1)+C

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integral of (15)/((0.5t+1)^2)

\int \frac{15}{( 0.5 t + 1 ) ^{2}} d t

- \frac{30}{0.5 t + 1} + C

Solution steps

\int \frac{15}{( 0.5 t + 1 ) ^{2}} d t

Take the constant out:

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

= 15 \cdot \int \frac{1}{( 0.5 t + 1 ) ^{2}} d t

Apply u-substitution

: \int \frac{1}{0.5 u ^{2}} d u

= 15 \cdot \int \frac{1}{0.5 u ^{2}} d u

Take the constant out:

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

= 15 \cdot \frac{1}{0.5} \cdot \int \frac{1}{u ^{2}} d u

Simplify

= 15 \cdot 2 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 15 \cdot 2 (- \frac{1}{u})

Substitute back

u = 0.5 t + 1

= 15 \cdot 2 (- \frac{1}{0.5 t + 1})

Simplify

15 \cdot 2 (- \frac{1}{0.5 t + 1}) : - \frac{30}{0.5 t + 1}

= - \frac{30}{0.5 t + 1}

Add a constant to the solution

= - \frac{30}{0.5 t + 1} + C

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What is the integral of (15)/((0.5t+1)^2) ?
The integral of (15)/((0.5t+1)^2) is -(30)/(0.5t+1)+C