If f(x) = int frac{1}{x^{1/4}(1+x^{1/4})} dx | vedclass

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(A) Let $x = t^4$,then $dx = 4t^3 dt$.Substituting these into the integral:$\int \frac{1}{t(1+t)} \cdot 4t^3 dt = \int \frac{4t^2}{1+t} dt$.Using poly

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$4(\log_e 2 - 2)$

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(A) Let $x = t^4$,then $dx = 4t^3 dt$.Substituting these into the integral:$\int \frac{1}{t(1+t)} \cdot 4t^3 dt = \int \frac{4t^2}{1+t} dt$.Using polynomial division or adjustment:$\int \frac{4(t^2-1+1)}{t+1} dt = 4 \int (t-1 + \frac{1}{t+1}) dt$.Integrating term by term:$f(x) = 4(\frac{t^2}{2} - t + \log_e|t+1|) + C = 2t^2 - 4t + 4\log_e(t+1) + C$.Substituting $t = x^{1/4}$:$f(x) = 2x^{1/2} - 4x^{1/4} + 4\log_e(1+x^{1/4}) + C$.Given $f(0) = -6$:$2(0) - 4(0) + 4\log_e(1) + C = -6 \implies C = -6$.Thus,$f(x) = 2x^{1/2} - 4x^{1/4} + 4\log_e(1+x^{1/4}) - 6$.Calculating $f(1)$:$f(1) = 2(1) - 4(1) + 4\log_e(2) - 6 = 2 - 4 + 4\log_e 2 - 6 = 4\log_e 2 - 8 = 4(\log_e 2 - 2)$.

If $f(x) = \int \frac{1}{x^{1/4}(1+x^{1/4})} dx$ and $f(0) = -6$,then $f(1)$ is equal to:

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