What is the integral of ((5t-2))/(t+1) ?

The integral of ((5t-2))/(t+1) is 5(t+1)-7ln|t+1|+C

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integral of ((5t-2))/(t+1)

\int \frac{( 5 t - 2 )}{t + 1} d t

5 (t + 1) - 7 ln ∣ t + 1 ∣ + C

Solution steps

\int \frac{5 t - 2}{t + 1} d t

Apply u-substitution

: \int \frac{5 u - 7}{u} d u

= \int \frac{5 u - 7}{u} d u

Expand

\frac{5 u - 7}{u} : 5 - \frac{7}{u}

= \int 5 - \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 5 d u - \int \frac{7}{u} d u

\int 5 d u = 5 u

\int \frac{7}{u} d u = 7 ln ∣ u ∣

= 5 u - 7 ln ∣ u ∣

Substitute back

u = t + 1

= 5 (t + 1) - 7 ln ∣ t + 1 ∣

Add a constant to the solution

= 5 (t + 1) - 7 ln ∣ t + 1 ∣ + C

\frac{d y}{d x} = \frac{( x ^{2} - 1 )}{y ^{2}}

\int \frac{e ^{x}}{3} - 4 x^{3} + \frac{6}{x} d x

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

n = 1 \sum \infty \frac{3}{n ^{2}}

(x y^{2} + x) d x + (y - x^{2} y) d y = 0

What is the integral of ((5t-2))/(t+1) ?
The integral of ((5t-2))/(t+1) is 5(t+1)-7ln|t+1|+C