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inverse of (x^2+1)/(x^2-x-6)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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∫₋₁²6x²+1 dx, n=3

Solution

11.44615…

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Solve by:

Solve by midpoint Riemann sum

Solve by left point Riemann sum

Solve by right point Riemann sum

Solve by trapezoidal rule

Solve by Simpson's rule

One step at a time

Riemann sum using midpoint of subinterval

∫_a^bf(x)dx ≈ Δ x(f(x₀+x₁2 )+f(x₁+x₂2 )+f(x₂+x₃2 )+...+f(x_n−1+x_n2 )),
where Δ x = b−an

Apply Riemann Formula

=1· (f(x₀+x₁2 )+f(x₁+x₂2 )+f(x₂+x₃2 ))

Calculate Sub Intervals

=1· (245 +245 +2413 )

=11.44615…

Long Arithmetic Examples

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

inverse of (x^2+1)/(x^2-x-6)

Solution

Solution steps

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