How do you find the integral of $\int \frac{x^{4} - 1}{x^{2} + 1} d x$ $int (x^4-1)/(x^2+1)dx$ ?

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How do you find the integral of $\int \frac{x^{4} - 1}{x^{2} + 1} d x$ $int (x^4-1)/(x^2+1)dx$ ?

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Answer 1 · 2016-12-04T15:46:02

We have that

$\int \frac{x^{4} - 1}{x^{2} + 1} d x = \int \frac{(x^{2} + 1) \cdot (x^{2} - 1)}{x^{2} + 1} d x = \int (x^{2} - 1) d x = \frac{x^{3}}{3} - x + c$ $int (x^4-1)/(x^2+1)dx=int [(x^2+1)*(x^2-1)]/(x^2+1)dx= int (x^2-1)dx=x^3/3-x+c$

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How do you find the integral of $\int \frac{x^{4} - 1}{x^{2} + 1} d x$ $int (x^4-1)/(x^2+1)dx$ ?

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How do you find the integral of $\int \frac{x^{4} - 1}{x^{2} + 1} d x$ $int (x^4-1)/(x^2+1)dx$ ?