How do solve the following integral? $\int \frac{{(t + 7)}^{2}}{t^{3}} d t$ $int(t+7)^2/t^3dt$

Question

How do solve the following integral? $\int \frac{{(t + 7)}^{2}}{t^{3}} d t$ $int(t+7)^2/t^3dt$

$\int \frac{{(t + 7)}^{2}}{t^{3}} d t$ $int(t+7)^2/t^3dt$

2 Answers

Impact of this question

2220 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-04T01:13:13

Just expand the function and integrate. The answer is $ln t - \frac{14}{t} - \frac{49}{2 t^{2}} + C$ $lnt-14/t-49/(2t^2)+C$ .

Explanation:

Let $f (t) = \frac{{(t + 7)}^{2}}{t^{3}}$ $f(t)=(t+7)^2/t^3$ . First, let's rearrange $f (t)$ $f(t)$ into simpler terms.
$f (t) = \frac{t^{2} + 14 t + 49}{t^{3}}$ $f(t)=(t^2+14t+49)/t^3$
$= \frac{1}{t} + \frac{14}{t^{2}} + \frac{49}{t^{3}}$ $=1/t+14/t^2+49/t^3$
$= t^{- 1} + 14 t^{- 2} + 49 t^{- 3}$ $=t^-1+14t^-2+49t^-3$

Then integrate $f (t)$ $f(t)$ . Note that $\int x^{n} d x = \frac{1}{n + 1} x^{n + 1} + C$ $intx^ndx=1/(n+1)x^(n+1)+C$ if $n \neq - 1$ $n!=-1$ , where $C$ $C$ is the integral constant. If $n = - 1$ $n=-1$ , $\int x^{- 1} d x = \int (\frac{1}{x}) d x = ln x + C$ $intx^-1dx=int(1/x)dx=lnx +C$ is applied.

$\int f (t) d t = \int (t^{- 1} + 14 t^{- 2} + 49 t^{- 3}) d t$ $intf(t)dt=int(t^-1+14t^-2+49t^-3)dt$
$= \int t^{- 1} d t + 14 \int t^{- 2} d t + 49 \int t^{- 3} d t$ $=intt^-1dt+14intt^-2dt+49intt^-3dt$
$= ln t + 14 (- t^{- 1}) + 49 (- \frac{1}{2} t^{- 2}) + C$ $=lnt+14(-t^-1)+49(-1/2t^-2)+C$
$= ln t - \frac{14}{t} - \frac{49}{2 t^{2}} + C$ $=lnt-14/t-49/(2t^2)+C$ .

Answer 2 · 2017-12-04T01:17:15

$ln (t) - \frac{7 (4 t + 7)}{2 t^{2}} + c$ $ln(t)-(7(4t+7))/(2t^2)+c$

Explanation:

solving ${(t + 7)}^{2}$ $(t+7)^2$

${(t + 7)}^{2} = t^{2} + 14 t + 49$ $(t+7)^2=t^2+14t+49$

now in the integral

$\int \frac{t^{2} + 14 t + 49}{t^{3}} d t$ $int (t^2+14t+49)/t^3 dt$

separating the fractions

$\int (\frac{t^{2}}{t^{3}} + 14 \frac{t}{t^{3}} + \frac{49}{t^{3}}) d t$ $int (t^2/t^3+14t/t^3+49/t^3) dt$

$\int \frac{1}{t} + \frac{14}{t^{2}} + \frac{49}{t^{3}} d t$ $int 1/t+14/t^2+49/t^3 dt$

separting

$\int \frac{1}{t} d t = ln (t) + c$ $int 1/tdt=ln(t)+c$

$\int \frac{14}{t^{2}} d t = - \frac{14}{t} + c$ $int 14/t^2dt=-14/t+c$

$\int \frac{49}{t^{3}} d t = - \frac{49}{2 t^{2}} + c$ $int49/t^3dt=-49/(2t^2)+c$

$\int \frac{1}{t} + \frac{14}{t^{2}} + \frac{49}{t^{3}} d t = ln (t) - \frac{14}{t} - \frac{49}{2 t^{2}} + c$ $int 1/t+14/t^2+49/t^3 dt=ln(t)-14/t-49/(2t^2)+c$

operating

$\int \frac{1}{t} + \frac{14}{t^{2}} + \frac{49}{t^{3}} d t = ln (t) - \frac{7 (4 t + 7)}{2 t^{2}} + c$ $int 1/t+14/t^2+49/t^3 dt=ln(t)-(7(4t+7))/(2t^2)+c$

$\int \frac{t^{2} + 14 t + 49}{t^{3}} d t = ln (t) - \frac{7 (4 t + 7)}{2 t^{2}} + c$ $int (t^2+14t+49)/t^3 dt=ln(t)-(7(4t+7))/(2t^2)+c$

Science

Math

Humanities

... and beyond

How do solve the following integral? $\int \frac{{(t + 7)}^{2}}{t^{3}} d t$ $int(t+7)^2/t^3dt$

$\int \frac{{(t + 7)}^{2}}{t^{3}} d t$ $int(t+7)^2/t^3dt$

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do solve the following integral? ∫(t+7)2t3dtint(t+7)^2/t^3dt

∫(t+7)2t3dtint(t+7)^2/t^3dt

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do solve the following integral? $\int \frac{{(t + 7)}^{2}}{t^{3}} d t$ $int(t+7)^2/t^3dt$

$\int \frac{{(t + 7)}^{2}}{t^{3}} d t$ $int(t+7)^2/t^3dt$