Integral 1/((x^2)-8x+17) dx
upper limit: Infinity
lower limit:4
upper limit: Infinity
lower limit:4
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The indefinite integral is arctan(-4 + x) + C
since the function is 1 / [(x-4)² + 1]
... and ∫ 1 / (X² +1) dX = arctan(X) +C
The value of the integral between x = 4 and x -> ∞ is π/2.
since the function is 1 / [(x-4)² + 1]
... and ∫ 1 / (X² +1) dX = arctan(X) +C
The value of the integral between x = 4 and x -> ∞ is π/2.
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π/2
Here's how. First complete the square n the denominator.
You get (x–4)² +1. So you should see immediately that it is an inverse tan function.
So the integral is tan‾¹(x–4) from x=4 to x=∞. The angle whose tangent is ∞ is π/2; the angle whose tanget is 0 is 0.
For full details about the integration, click Show Steps here:
http://www.wolframalpha.com/input/?i=Int…
Here's how. First complete the square n the denominator.
You get (x–4)² +1. So you should see immediately that it is an inverse tan function.
So the integral is tan‾¹(x–4) from x=4 to x=∞. The angle whose tangent is ∞ is π/2; the angle whose tanget is 0 is 0.
For full details about the integration, click Show Steps here:
http://www.wolframalpha.com/input/?i=Int…