How do you integrate $\int {(1 + \frac{1}{t})}^{3} \cdot \frac{1}{t^{2}}$ $int (1+1/t)^3*1/t^2$ using substitution?

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Answer 1 · 2016-07-21T23:20:33

$\int {(1 + \frac{1}{t})}^{3} \cdot \frac{1}{t^{2}} d t = - \frac{{(t + 1)}^{4}}{4 t} + C$ $int (1+1/t)^3 * 1/t^2 dt =-(t+1)^4/(4t) + C$

Let $u = 1 + \frac{1}{t} \Rightarrow d u = - \frac{1}{t^{2}} d t$ $u = 1 + 1/t implies du = -1/t^2dt$

$\int {(1 + \frac{1}{t})}^{3} \cdot \frac{1}{t^{2}} d t = - \int u^{3} d u$ $int (1+1/t)^3 * 1/t^2 dt = - int u^3du$

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

Tidying up a little gives:

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

How do you integrate ∫(1+1t)3⋅1t2int (1+1/t)^3*1/t^2 using substitution?