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

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

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

\int \frac{( 4 x - 3 )}{x + 1} d x

4 (x + 1) - 7 ln ∣ x + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

= \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 = x + 1

= 4 (x + 1) - 7 ln ∣ x + 1 ∣

Add a constant to the solution

= 4 (x + 1) - 7 ln ∣ x + 1 ∣ + C

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

f (x) = 2 x^{2} + 3

\frac{d}{d x} (x^{(\frac{1}{3})})

f (x) = 10 x^{2} - 56 x + 104

\frac{d}{d x} (- \frac{1}{y - x})

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