The value of int frac{x^3}{sqrt{1 + x^4}} , d | vedclass

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(C) To solve the integral $I = \int \frac{x^3}{\sqrt{1 + x^4}} \, dx$,we use the method of substitution.Let $t = 1 + x^4$.Then,differentiating both si

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$\frac{1}{2}(1 + x^4)^{1/2} + c$

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(C) To solve the integral $I = \int \frac{x^3}{\sqrt{1 + x^4}} \, dx$,we use the method of substitution.Let $t = 1 + x^4$.Then,differentiating both sides with respect to $x$,we get $dt = 4x^3 \, dx$,which implies $x^3 \, dx = \frac{1}{4} \, dt$.Substituting these into the integral:$I = \int \frac{1}{\sqrt{t}} \cdot \frac{1}{4} \, dt = \frac{1}{4} \int t^{-1/2} \, dt$.Applying the power rule for integration $\int t^n \, dt = \frac{t^{n+1}}{n+1} + c$:$I = \frac{1}{4} \cdot \frac{t^{1/2}}{1/2} + c = \frac{1}{4} \cdot 2 \cdot t^{1/2} + c = \frac{1}{2} \sqrt{t} + c$.Substituting back $t = 1 + x^4$,we get:$I = \frac{1}{2} \sqrt{1 + x^4} + c$.

The value of $\int \frac{x^3}{\sqrt{1 + x^4}} \, dx$ is

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