If int {frac{1}{{x + {x^5}}}dx = f(x) + c} ,t | vedclass

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(A) Let $I = \int \frac{x^4}{x + x^5} dx$.We can rewrite the numerator as $(x^4 + 1) - 1$.$I = \int \frac{x^4 + 1 - 1}{x(1 + x^4)} dx$$I = \int \frac{

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$\log |x| - f(x) + c$

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(A) Let $I = \int \frac{x^4}{x + x^5} dx$.We can rewrite the numerator as $(x^4 + 1) - 1$.$I = \int \frac{x^4 + 1 - 1}{x(1 + x^4)} dx$$I = \int \frac{x^4 + 1}{x(1 + x^4)} dx - \int \frac{1}{x(1 + x^4)} dx$$I = \int \frac{1}{x} dx - \int \frac{1}{x + x^5} dx$Given that $\int \frac{1}{x + x^5} dx = f(x) + c_1$,we substitute this into the expression:$I = \log |x| - (f(x) + c_1) + c_2$$I = \log |x| - f(x) + c$,where $c = c_2 - c_1$ is a new constant.Thus,the correct option is $A$.

If $\int {\frac{1}{{x + {x^5}}}dx = f(x) + c} $,then the value of $\int {\frac{{{x^4}}}{{x + {x^5}}}dx} $ is

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