Which diagram correctly shows the relationship

Ответ:

hannahbeccahxo9681

08.09.2022

Answer: ∫(x²+1)/(x⁴-1)dx is equal to $-\frac{1}{2} \ln|x+1| + \frac{1}{2}\ln|x-1| + c$ ====== $\displaystyle\int \frac{x^2+1}{x^4-1} dx$ note that, by difference of squares, x⁴-1 = (x²+1)(x²-1), so $\displaystyle\int \frac{x^2+1}{x^4-1} dx \\ \\ = \displaystyle\int \frac{x^2+1}{(x^2+1)(x^2-1)} dx$ (x²+1) cancel so we are left with $= \displaystyle\int \frac{1}{x^2-1} dx \\ \\ = \int \frac{1}{(x+1)(x-1)} dx$ we can break down $\frac{1}{(x+1)(x-1)}$ into partial fractionssince there are two linear factors in the denominator, x + 1 and x - 1, we have the following partial fraction decomposition: $\displaystyle \frac{1}{(x+1)(x-1)} = \frac{a}{x+1} + \frac{b}{x-1}$ we are trying to make an identity because we want to change the form of $\frac{1}{(x+1)(x-1)}$ into a decomposed partial fraction form. therefore, this equation is true for all values of x.to find the values of a and b, we can start by multiplying both sides of the equation by the lowest common denominator, (x+1)(x-1). $\begin{aligned} (x+1)(x-1) \cdot \frac{1}{(x+1)(x-1)} & = \left(\frac{a}{x+1} + \frac{b}{x-1}\right) \cdot (x+1)(x-1) \\ 1 & = a(x-1) + b(x+1) \end{aligned}$ we can use a shortcut for linear factors. since this equation is an identity and true for all values of x, we can use any value of x to us try to find the values of a and b that will establish this identity.if we choose x = 1, we can eliminate a and solve for b: $\begin{aligned} 1 & = a(1-1) + b(1+1) \\ 1 & = 0 + 2b \\ b & = 1/2 \end{aligned}$ if we choose x = -1, we can eliminate b and solve for a $\begin{aligned} 1 & = a(-1-1) + b(-1+1) \\ 1 & = -2a + 0 \\ a & = -1/2 \end{aligned}$ therefore: $\begin{aligned} \frac{1}{(x+1)(x-1)} & = \frac{-\frac{1}{2}}{x+1} + \frac{\frac{1}{2}}{x-1}\end{aligned}$ and we have $\begin{aligned} \displaystyle & = \int \frac{1}{(x+1)(x-1)} dx \\ \\ & = \int \left(\frac{-\frac{1}{2}}{x+1} + \frac{\frac{1}{2}}{x-1} \right) dx \\ \\ & = \int \frac{-\frac{1}{2}}{x+1} dx + \int \frac{\frac{1}{2}}{x-1}dx & & \text{\footnotesize (integral sum rule)} \\ \\ & = -\frac{1}{2}\int \frac{1}{x+1}dx + \frac{1}{2}\int \frac{1}{x-1}dx& & \text{\footnotesize (integral constant rule)} \\ \\ & = -\frac{1}{2} \ln|x+1| + \frac{1}{2}\ln|x-1| + c & & \text{\footnotesize ($u$-substitution)} \end{aligned}$

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