What is the integral of (6t-5)/(t+1) ?

The integral of (6t-5)/(t+1) is 6(t+1)-11ln|t+1|+C

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integral of (6t-5)/(t+1)

\int \frac{6 t - 5}{t + 1} d t

6 (t + 1) - 11 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{6 t - 5}{t + 1} d t

Apply u-substitution

: \int \frac{6 u - 11}{u} d u

= \int \frac{6 u - 11}{u} d u

Expand

\frac{6 u - 11}{u} : 6 - \frac{11}{u}

= \int 6 - \frac{11}{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 6 d u - \int \frac{11}{u} d u

\int 6 d u = 6 u

\int \frac{11}{u} d u = 11 ln ∣ u ∣

= 6 u - 11 ln ∣ u ∣

Substitute back

u = t + 1

= 6 (t + 1) - 11 ln ∣ t + 1 ∣

Add a constant to the solution

= 6 (t + 1) - 11 ln ∣ t + 1 ∣ + C

f (x) = 5 x^{3}

(1 + x) d y = (y + x) d x

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

\int_{- π}^{0} 0 d x

\int_{0}^{4} \frac{1}{( 9 - 3 x ) ^{\frac{1}{3}}} d x

What is the integral of (6t-5)/(t+1) ?
The integral of (6t-5)/(t+1) is 6(t+1)-11ln|t+1|+C