int {{x^3}sqrt {3 + 5{x^4}} } ;dx = | vedclass

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Question

(C) Let $I = \int {{x^3}\sqrt {3 + 5{x^4}} } \;dx$. Substitute $3 + 5{x^4} = t$. Differentiating both sides with respect to $x$,we get $20{x^3}dx = dt

Vedclass · Accepted Answer

$\frac{1}{30}{(3 + 5{x^4})^{3/2}} + C$

Vedclass · Answer

(C) Let $I = \int {{x^3}\sqrt {3 + 5{x^4}} } \;dx$. Substitute $3 + 5{x^4} = t$. Differentiating both sides with respect to $x$,we get $20{x^3}dx = dt$,which implies ${x^3}dx = \frac{1}{20}dt$. Substituting these into the integral: $I = \int \sqrt{t} \cdot \frac{1}{20} dt = \frac{1}{20} \int t^{1/2} dt$. Using the power rule $\int t^n dt = \frac{t^{n+1}}{n+1} + C$: $I = \frac{1}{20} \cdot \frac{t^{3/2}}{3/2} + C = \frac{1}{20} \cdot \frac{2}{3} t^{3/2} + C = \frac{1}{30} t^{3/2} + C$. Substituting back $t = 3 + 5{x^4}$,we get: $I = \frac{1}{30}{(3 + 5{x^4})^{3/2}} + C$.

$\int {{x^3}\sqrt {3 + 5{x^4}} } \;dx = $

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