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

The integral of (1+1/t)^3(1/(t^2)) is -((t+1)^4)/(4t^4)+C

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

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

- \frac{( t + 1 ) ^{4}}{4 t ^{4}} + C

Solution steps

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

Apply u-substitution

: \int - u^{3} d u

= \int - u^{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= - \frac{u ^{4}}{4}

Substitute back

u = 1 + \frac{1}{t}

= - \frac{( 1 + \frac{1}{t} ) ^{4}}{4}

Simplify

- \frac{( 1 + \frac{1}{t} ) ^{4}}{4} : - \frac{( t + 1 ) ^{4}}{4 t ^{4}}

= - \frac{( t + 1 ) ^{4}}{4 t ^{4}}

Add a constant to the solution

= - \frac{( t + 1 ) ^{4}}{4 t ^{4}} + C

n = 1 \sum \infty (0.1)^{n}

\frac{d}{d x} (\frac{2 x + 5}{x ^{2} - 3})

f (x) = - 3 x + 4

(0, 0), (3, 4)

- \frac{3}{4 s ^{4}} + \frac{10}{3 s ^{3}}

What is the integral of (1+1/t)^3(1/(t^2)) ?
The integral of (1+1/t)^3(1/(t^2)) is -((t+1)^4)/(4t^4)+C