What is the integral of 5/(t+1) ?

The integral of 5/(t+1) is 5ln|t+1|+C

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integral of 5/(t+1)

\int \frac{5}{t + 1} d t

5 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{5}{t + 1} d t

Take the constant out:

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

= 5 \cdot \int \frac{1}{t + 1} d t

Apply u-substitution

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

= 5 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 ln ∣ u ∣

Substitute back

u = t + 1

= 5 ln ∣ t + 1 ∣

Add a constant to the solution

= 5 ln ∣ t + 1 ∣ + C

\int \frac{1}{16 - 25 x ^{2}} d x

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\int (4 x^{2} + 8 x + 9)^{6} (x + 1) d x

arccos (x)

x \to 0 lim (\frac{e ^{x}}{x ^{2}})