Best solutions to - "Find the following integral. ∫(t+3)2t4dt"

miyoko23q3

Answered question

2021-11-07

Find the following integral.

\int \frac{{(t + 3)}^{2}}{t^{4}} d t

George Spencer

Beginner2021-11-08Added 12 answers

Step 1
We have to find the integral:

\int \frac{{(t + 3)}^{2}}{t^{4}} d t

Simplifying using identity

{(a + b)}^{2} = a^{2} + 2 a b + b^{2}

\int \frac{{(t + 3)}^{2}}{t^{4}} d t = \int \frac{t^{2} + 2 (t) (3) + 3^{2}}{t^{4}} d t

= \int \frac{t^{2} + 6 t + 9}{t^{4}} d t

Separating each terms,

\int \frac{t^{2} + 6 t + 9}{t^{4}} d t = \int (\frac{t^{2}}{t^{4}} + \frac{6 t}{t^{4}} + \frac{9}{t^{4}}) d t

= \int (\frac{1}{t^{2}} + \frac{6}{t^{3}} + \frac{9}{t^{4}}) d t

= \int t^{- 2} d t + 6 \int t^{- 3} d t + 9 \int t^{- 4} d t

Step 2
We know the formula of integration,

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

Where, C is an arbitrary constant.
So integral would be following,

\int t^{- 2} d t + 6 \int t^{- 3} d t + 9 \int t^{- 4} d t = \frac{t^{- 2 + 1}}{- 2 + 1} + 6 \frac{t^{- 3 + 1}}{- 3 + 1} + 9 \frac{t^{- 4 + 1}}{- 4 + 1} + C

= \frac{t^{- 1}}{- 1} + 6 \frac{t^{- 2}}{- 2} + 9 \frac{t^{- 3}}{- 3} + C

= - \frac{1}{t} - \frac{3}{t^{2}} - \frac{3}{t^{3}} + C

Hence, value of given integral is

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