# Evaluate the following indefinite integral. `int((11+4sqrt(t))^2)/(t^(1/3))dt` integrate(((11+4sqrt(t))^2)/(t^(1/3)))

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use the following substitution such that:

`4sqrt t = u - 11 => sqrt t = (u-11)/4 => t = (u-11)^2/16`

`4/(2sqrt t) dt = du => 2/sqrt t dt = du => dt = du*sqrt t/2`

`dt = (u-11)/8 du`

Changing the variable yields:

`int (11+4sqrt t)^2/(t^(1/3)) dt = int (u^2(u-11)root(3)(16))/(8(u-11)^(2/3)) du` = `2root(3)(2)/8 int u^2(u-11)^(1/3)du`

You should perform one more substitution such that:

`u - 11 = v => du = dv`

`u = 11 + v => u^2 = (11 + v)^2`

Changing the variable yields:

`root(3)(2)/4 int (11 + v)^2*v^(1/3) dv`

Expanding the square yields:

`root(3)(2)/4 int (121 + 22v + v^2)*v^(1/3) dv`

`root(3)(2)/4 (int 121v^(1/3) dv + int 22v*v^(1/3) dv + int v^2*v^(1/3) dv)`

`root(3)(2)/4 (121(v^(1/3+1))/(1/3+1) + 22(v^(1/3+2))/(1/3+2) +(v^(1/3+3))/(1/3+3))` + c

`root(3)(2)/4 (363(v^(4/3))/4 + 66(v^(7/3))/7 +(3v^(10/3))/10)` + c

`root(3)(2)/4 (363((4sqrt t)^(4/3))/4 + 66((4 sqrt t)^(7/3))/7 +(3(4 sqr t)^(10/3))/10)` + c

Hence, evaluating the given integral yields `int (11+4sqrt t)^2/(t^(1/3)) dt =root(3)(2)/4 (363((4sqrt t)^(4/3))/4 + 66((4 sqrt t)^(7/3))/7 +((6 sqrt t)^(10/3))/5) + c.`

Sources:

lemjay | High School Teacher | (Level 3) Senior Educator

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`int (11+4sqrtt)^2/t^(1/3) dt`

To integrate, let's use integration by parts. The formula is `intudv=uv-intvdu` .

Let,

`u=(11+4sqrtt)^2`                            and           `dv=1/t^(1/3)dt `

`du=2(11+4sqrtt)(11+4sqrtt)'dt`                      `dv= t^(-1/3)dt`

`du=2(11+4sqrtt)(2/sqrtt)dt `                                 `v = int t^(-1/3)dt`

`du= (4(11+4sqrtt))/sqrtt dt`                                           `v = 3/2t^(2/3)`

Susbtitute u,v and du to the formula of integration by parts.

`int(11+4sqrtt)^2/t^(1/3)dt = (11+4sqrtt)^2*3/2t^(2/3) - int 3/2t^(2/3)*(4(11+4sqrtt))/(sqrtt)dt`

`=3/2(11+4sqrtt)t^(2/3) - 6int t^(2/3)*(11+4t^(1/2))/t^(1/2) dt`

`=3/2(11+4sqrtt)t^(2/3) - 6 int t^(1/6)(11+4t^(1/2))dt`

`= 3/2(11+4sqrtt)t^(2/3)-6int(11t^(1/6)+4t^(2/3))dt`

`=3/2(11+4sqrtt)t^(2/3) -6[ int11t^(1/6)dt + int4t^(2/3)dt]`

`=3/2(11+4sqrtt)t^(2/3) - 6[66/7t^(7/6) + 12/5t^(5/3)]`

`=33/2t^(2/3)+6t^(7/6) - 396/7t^(7/6) - 72/5t^(5/3)`

`=33/2t^(2/3) - 72/5t^(5/3)-354/7t^(7/6)`

--------------------------------------------------------------------------------------         Since we have indefinite integral, therefore

`int(11+4sqrtt)^2/t^(1/3)dt = 33/2t^(2/3)-42/5t^(5/3)-396/7t^(7/6) + C` .