What is the integral of (7t-2)/(t+1) ?

The integral of (7t-2)/(t+1) is 7(t+1)-9ln|t+1|+C

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integral of (7t-2)/(t+1)

\int \frac{7 t - 2}{t + 1} d t

7 (t + 1) - 9 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{7 t - 2}{t + 1} d t

Apply u-substitution

: \int \frac{7 u - 9}{u} d u

= \int \frac{7 u - 9}{u} d u

Expand

\frac{7 u - 9}{u} : 7 - \frac{9}{u}

= \int 7 - \frac{9}{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{9}{u} d u

\int 7 d u = 7 u

\int \frac{9}{u} d u = 9 ln ∣ u ∣

= 7 u - 9 ln ∣ u ∣

Substitute back

u = t + 1

= 7 (t + 1) - 9 ln ∣ t + 1 ∣

Add a constant to the solution

= 7 (t + 1) - 9 ln ∣ t + 1 ∣ + C

\int (\frac{x ^{2}}{x ^{2} + 1}) d x

\int \frac{1}{( x - 6 ) ( x - 3 ) ( x + 1 )} d x

y^{^{'}} = 100 x - x y

\frac{\partial}{\partial x} (- 9 x^{2} y^{2} - cos (y))

x^{3} + 8

What is the integral of (7t-2)/(t+1) ?
The integral of (7t-2)/(t+1) is 7(t+1)-9ln|t+1|+C