What is the integral of (5x-3)/(x+1) ?

The integral of (5x-3)/(x+1) is 5(x+1)-8ln|x+1|+C

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integral of (5x-3)/(x+1)

\int \frac{5 x - 3}{x + 1} d x

5 (x + 1) - 8 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{5 x - 3}{x + 1} d x

Apply u-substitution

: \int \frac{5 u - 8}{u} d u

= \int \frac{5 u - 8}{u} d u

Expand

\frac{5 u - 8}{u} : 5 - \frac{8}{u}

= \int 5 - \frac{8}{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 5 d u - \int \frac{8}{u} d u

\int 5 d u = 5 u

\int \frac{8}{u} d u = 8 ln ∣ u ∣

= 5 u - 8 ln ∣ u ∣

Substitute back

u = x + 1

= 5 (x + 1) - 8 ln ∣ x + 1 ∣

Add a constant to the solution

= 5 (x + 1) - 8 ln ∣ x + 1 ∣ + C

f (x) = \frac{4}{3 x ^{2}}

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

\int - \frac{18}{x ^{19}} d x

\frac{d x}{d t} - \frac{x}{t - 1} = t^{2} + 2

x \to 0 lim (\frac{( 1 + x ) ^{2} - 1}{x})

What is the integral of (5x-3)/(x+1) ?
The integral of (5x-3)/(x+1) is 5(x+1)-8ln|x+1|+C