What is the integral of (8t-3)/(t+1) ?

The integral of (8t-3)/(t+1) is 8(t+1)-11ln|t+1|+C

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integral of (8t-3)/(t+1)

\int \frac{8 t - 3}{t + 1} d t

8 (t + 1) - 11 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{8 t - 3}{t + 1} d t

Apply u-substitution

: \int \frac{8 u - 11}{u} d u

= \int \frac{8 u - 11}{u} d u

Expand

\frac{8 u - 11}{u} : 8 - \frac{11}{u}

= \int 8 - \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 8 d u - \int \frac{11}{u} d u

\int 8 d u = 8 u

\int \frac{11}{u} d u = 11 ln ∣ u ∣

= 8 u - 11 ln ∣ u ∣

Substitute back

u = t + 1

= 8 (t + 1) - 11 ln ∣ t + 1 ∣

Add a constant to the solution

= 8 (t + 1) - 11 ln ∣ t + 1 ∣ + C

x \to 4 lim (x^{2} + 5 - x)

t^{2} y^{^{''}} + 3 t y^{^{'}} - 8 y = 0

\frac{\partial}{\partial y} (x^{4} + x^{2} y)

\frac{d}{d x} (\frac{x ^{2} + 3 x}{6})

\int_{0}^{16} x - 4 d x

What is the integral of (8t-3)/(t+1) ?
The integral of (8t-3)/(t+1) is 8(t+1)-11ln|t+1|+C