Question

evaluate the integral of

dx/square root of 9-8x-x^2
9-8x-x^2 = 25 - (16 + 8x + x^2) = 3^2 - (x-4)^2

Integral sqrt(a^2 - u^2) du = u/2 * sqrt(a^2 - u^2) + a^2/2 arcsin(u/a)

just plug in a=5 and u=x-4 and plug away.

This is a standard trig substitution and integration by parts.
typo: 5^2 not 3^2
Go on: wolframalpha dot com

When page be open in rectangle type:

integrate dx/sqrt(9-8x-x^2)

and click option =

After few seconds when you see result click option: Show steps

Related Questions

how do i evaluate the integral of dx/ x^2 +9 evaluate the integral: (x^3)/((x^2)+1) Evaluate Integral x^2+4 / x^4+x^2+16 dx Evaluate Integral of (lnx)^4/x dx when b is E and a is 1 Evaluate the integral Int 7x ln(5x) dx evaluate the integral from 4 to 3. x/2x^2-6dx Evaluate the integral. ∫ 28e^(√7x)/(2√x) dx Evaluate the integral. ∫5sec^(4)x dx I got 5(secx+tanx)^(5)+C. is it right. Evaluate the integral of (e^2x)*sin^3 x dx I let u = e^2x, du = (1/2)e^2x dx v= (-1/3)cos^... How can I evaluate the integral of xln(x)dx by using integration by parts