How do you integrate by substitution $\int (\frac{t^{3}}{3} + \frac{1}{4 t^{2}}) d t$ $int (t^3/3+1/(4t^2))dt$ ?

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Answer 1 · 2018-06-19T18:45:45

$= \frac{t^{4}}{12} - \frac{1}{4 t} + c$ $=t^4/12-1/(4t)+c$

you don't do this by substitution just use the power rule

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1}, n \neq - 1$ $intx^ndx=x^(n+1)/(n+1), n!=-1$

$\int (\frac{t^{3}}{3} + \frac{1}{4 t^{2}}) d t$ $int(t^3/3+1/(4t^2))dt$

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

$= \frac{1}{3} \frac{t^{4}}{4} + (\frac{1}{4}) \frac{t^{- 1}}{- 1} + c$ $=1/3(t^4)/4+(1/4)t^(-1)/(-1)+c$

$= \frac{t^{4}}{12} - \frac{1}{4 t} + c$ $=t^4/12-1/(4t)+c$

Answer 2 · 2018-06-19T18:48:20

$\frac{t^{4}}{12} - \frac{1}{4 t} + C$ $t^4/12-1/(4t)+C$

Writing
$\int (\frac{t^{3}}{3} + \frac{1}{4} t^{- 2}) d t$ $int( t^3/3+1/4t^(-2))dt$
we get

$\frac{t^{4}}{12} - \frac{1}{4 t} + C$ $t^4/12-1/(4t)+C$
we are using that

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $int x^ndx=x^(n+1)/(n+1)+C$ if $n \neq - 1$ $n ne -1$

How do you integrate by substitution int (t^3/3+1/(4t^2))dt∫(t33+14t2)dtint (t^3/3+1/(4t^2))dt?