How do you integrate $\int \frac{x^{4}}{x^{5} + 1}$ $int (x^4)/(x^(5)+1)$ ?

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How do you integrate $\int \frac{x^{4}}{x^{5} + 1}$ $int (x^4)/(x^(5)+1)$ ?

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Answer 1 · 2015-05-05T18:29:51

$\int \frac{x^{4} d x}{x^{5} + 1} = \frac{1}{5} ln (x^{5} + 1) + C$ $int(x^4dx)/(x^5+1)=1/5ln(x^5+1) +C$

Method

$\int \frac{x^{4} d x}{x^{5} + 1} = \frac{1}{5} \int \frac{5 x^{4} d x}{x^{5} + 1} = \frac{1}{5} \int \frac{d (x^{5} + 1)}{x^{5} + 1} = I$ $int(x^4dx)/(x^5+1)= 1/5int(5x^4dx)/(x^5+1)= 1/5int(d(x^5+1))/(x^5+1)= " I"$

NB: ** $d (x^{5} + 1)$ $d(x^5+1)$ simply means that the derivative of $x^{5} + 1$ $x^5+1$ is** $5 x^{4}$ $5x^4$

Now, you could as well let $u = x^{5} + 1$ $u=x^5+1$

So that,

$I = \frac{1}{5} \int \frac{d u}{u} = \frac{1}{5} ln u = \frac{1}{5} ln (x^{5} + 1) + C$ $"I"=1/5int(du)/u=1/5lnu=1/5ln(x^5+1) +C$

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How do you integrate $\int \frac{x^{4}}{x^{5} + 1}$ $int (x^4)/(x^(5)+1)$ ?

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