What is the integral of (4t-3)/(t+1) ?

The integral of (4t-3)/(t+1) is 4(t+1)-7ln|t+1|+C

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integral of (4t-3)/(t+1)

\int \frac{4 t - 3}{t + 1} d t

4 (t + 1) - 7 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{4 t - 3}{t + 1} d t

Apply u-substitution

= \int \frac{4 u - 7}{u} d u

Expand

\frac{4 u - 7}{u} : 4 - \frac{7}{u}

= \int 4 - \frac{7}{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 4 d u - \int \frac{7}{u} d u

\int 4 d u = 4 u

\int \frac{7}{u} d u = 7 ln ∣ u ∣

= 4 u - 7 ln ∣ u ∣

Substitute back

u = t + 1

= 4 (t + 1) - 7 ln ∣ t + 1 ∣

Add a constant to the solution

= 4 (t + 1) - 7 ln ∣ t + 1 ∣ + C

What is the integral of (4t-3)/(t+1) ?
The integral of (4t-3)/(t+1) is 4(t+1)-7ln|t+1|+C