What is the integral of ((t^2+t+1)/((t+3)^3)) ?

The integral of ((t^2+t+1)/((t+3)^3)) is -(t^2)/(2(t+3)^2)+ln|t+3|+2/(t+3)+1/((t+3)^2)+C

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integral of ((t^2+t+1)/((t+3)^3))

\int (\frac{t ^{2} + t + 1}{( t + 3 ) ^{3}}) d t

- \frac{t ^{2}}{2 ( t + 3 ) ^{2}} + ln ∣ t + 3 ∣ + \frac{2}{t + 3} + \frac{1}{( t + 3 ) ^{2}} + C

Solution steps

\int \frac{t ^{2} + t + 1}{( t + 3 ) ^{3}} d t

Expand

\frac{t ^{2} + t + 1}{( t + 3 ) ^{3}} : \frac{t ^{2}}{( t + 3 ) ^{3}} + \frac{t}{( t + 3 ) ^{3}} + \frac{1}{( t + 3 ) ^{3}}

= \int \frac{t ^{2}}{( t + 3 ) ^{3}} + \frac{t}{( t + 3 ) ^{3}} + \frac{1}{( t + 3 ) ^{3}} 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{t ^{2}}{( t + 3 ) ^{3}} d t + \int \frac{t}{( t + 3 ) ^{3}} d t + \int \frac{1}{( t + 3 ) ^{3}} d t

\int \frac{t ^{2}}{( t + 3 ) ^{3}} d t = - \frac{t ^{2}}{2 ( t + 3 ) ^{2}} + ln ∣ t + 3 ∣ + \frac{3}{t + 3}

\int \frac{t}{( t + 3 ) ^{3}} d t = - \frac{1}{t + 3} + \frac{3}{2 ( t + 3 ) ^{2}}

\int \frac{1}{( t + 3 ) ^{3}} d t = - \frac{1}{2 ( t + 3 ) ^{2}}

= - \frac{t ^{2}}{2 ( t + 3 ) ^{2}} + ln ∣ t + 3 ∣ + \frac{3}{t + 3} - \frac{1}{t + 3} + \frac{3}{2 ( t + 3 ) ^{2}} - \frac{1}{2 ( t + 3 ) ^{2}}

Simplify

- \frac{t ^{2}}{2 ( t + 3 ) ^{2}} + ln ∣ t + 3 ∣ + \frac{3}{t + 3} - \frac{1}{t + 3} + \frac{3}{2 ( t + 3 ) ^{2}} - \frac{1}{2 ( t + 3 ) ^{2}} : - \frac{t ^{2}}{2 ( t + 3 ) ^{2}} + ln ∣ t + 3 ∣ + \frac{2}{t + 3} + \frac{1}{( t + 3 ) ^{2}}

= - \frac{t ^{2}}{2 ( t + 3 ) ^{2}} + ln ∣ t + 3 ∣ + \frac{2}{t + 3} + \frac{1}{( t + 3 ) ^{2}}

Add a constant to the solution

= - \frac{t ^{2}}{2 ( t + 3 ) ^{2}} + ln ∣ t + 3 ∣ + \frac{2}{t + 3} + \frac{1}{( t + 3 ) ^{2}} + C

- y + 10 e^{- t} = y^{^{'}}

\int ln (6 x) d x

sin (2 t) + cos (2 t)

\int sin^{3} (3 θ) cos (3 θ) d θ

x y^{^{'}} + y = - 2 x y^{2}, y (1) = - 3

What is the integral of ((t^2+t+1)/((t+3)^3)) ?
The integral of ((t^2+t+1)/((t+3)^3)) is -(t^2)/(2(t+3)^2)+ln|t+3|+2/(t+3)+1/((t+3)^2)+C