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

The integral of 8/(t+1)+t is 8ln|t+1|+(t^2)/2+C

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integral of 8/(t+1)+t

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

8 ln ∣ t + 1 ∣ + \frac{t ^{2}}{2} + C

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{8}{t + 1} d t + \int t d t

\int \frac{8}{t + 1} d t = 8 ln ∣ t + 1 ∣

\int t d t = \frac{t ^{2}}{2}

= 8 ln ∣ t + 1 ∣ + \frac{t ^{2}}{2}

Add a constant to the solution

= 8 ln ∣ t + 1 ∣ + \frac{t ^{2}}{2} + C

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

\frac{\partial}{\partial x} (- e^{- x} y - e^{- x})

f (x) = (\frac{480 t}{( t ^{2} + 5 ) ^{3}})

\int \frac{x ( y ^{2} - 5 y + 6 )}{x ^{2} + 1} d x

x \to 5 lim (in (x^{2} - 25))

What is the integral of 8/(t+1)+t ?
The integral of 8/(t+1)+t is 8ln|t+1|+(t^2)/2+C