What is the integral of 3/(60-t) ?

The integral of 3/(60-t) is -3ln|60-t|+C

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integral of 3/(60-t)

\int \frac{3}{60 - t} d t

- 3 ln ∣ 60 - t ∣ + C

Solution steps

\int \frac{3}{60 - t} d t

Take the constant out:

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

= 3 \cdot \int \frac{1}{60 - t} d t

Apply u-substitution

: \int - \frac{1}{u} d u

= 3 \cdot \int - \frac{1}{u} d u

Take the constant out:

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

= 3 (- \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- ln ∣ u ∣)

Substitute back

u = 60 - t

= 3 (- ln ∣ 60 - t ∣)

Simplify

= - 3 ln ∣ 60 - t ∣

Add a constant to the solution

= - 3 ln ∣ 60 - t ∣ + C

\frac{8}{x ^{2} + 4}

f (x) = \frac{20 x}{x ^{2} - 5}

x \to - 6 + lim (\frac{1}{( x + 6 ) ^{6}})

\int_{0}^{3} - 6 x^{2} d x

f (x) = 2 x^{3} - 5 x + 4, at x = - 1