`int t^3 sqrt(t^4+1) dt` Find the indefinite integral

`int t^3 sqrt(t^4 +1) dt`

To solve, apply u-substitution method.

`u=t^4 + 1`

`du = 4t^3dt`

`(du)/4=t^3dt`

Expressing the integral in terms of u, it becomes:

`= int sqrt(t^4+1)*t^3dt`

`= int sqrtu * (du)/4`

`=1/4int sqrtu du`

Then, express the radical in exponent form.

`= 1/4 int u^(1/2)du`

To take the integral of this, apply the formula `int x^ndx = x^(n+1)/(n+1)+C`.

`=1/4*u^(3/2)/(3/2) + C`

`=1/4*u^(3/2)*2/3+C`

`=1/6u^(3/2)+C`

And, substitute back `u = t^4+1` .

`=1/6(t^4+1)^(3/2) + C`

Therefore, `int t^3sqrt(t^4+1)dt=1/6(t^4+1)^(3/2) + C` .

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