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

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

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

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

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

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

x \to 0 lim (cos (3 x))

\int x \cdot ab \cdot e^{- ab x + ab c} d x

\int 3 x^{2} sin (π x) d x

\frac{d}{d y} (\frac{y}{5})

t \to 0 lim (\frac{t ^{2} + 3 t}{t})

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