What is the integral of (7t-3)/(t+1) ?

The integral of (7t-3)/(t+1) is 7(t+1)-10ln|t+1|+C

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integral of (7t-3)/(t+1)

\int \frac{7 t - 3}{t + 1} d t

7 (t + 1) - 10 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{7 t - 3}{t + 1} d t

Apply u-substitution

: \int \frac{7 u - 10}{u} d u

= \int \frac{7 u - 10}{u} d u

Expand

\frac{7 u - 10}{u} : 7 - \frac{10}{u}

= \int 7 - \frac{10}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 7 d u - \int \frac{10}{u} d u

\int 7 d u = 7 u

\int \frac{10}{u} d u = 10 ln ∣ u ∣

= 7 u - 10 ln ∣ u ∣

Substitute back

u = t + 1

= 7 (t + 1) - 10 ln ∣ t + 1 ∣

Add a constant to the solution

= 7 (t + 1) - 10 ln ∣ t + 1 ∣ + C

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What is the integral of (7t-3)/(t+1) ?
The integral of (7t-3)/(t+1) is 7(t+1)-10ln|t+1|+C