Integral of 1/sqrt(polynomial) -- Outside the box thinking

See www.midnighttutor.com for all our calculus tutorials. Solution to integral of (1-2x)/sqrt(1-3x-2x^2) dx using complete the square, inverse trig functions, and other tricks.

Category:

Howto & Style

Tags:

• calculus
• integration
• integral
• inverse
• trig
• trigonometric
• complete
• square

[ElliotTM92]

@ksonggg Because that's not the actual rule. You're thinking of just ln(x).

But the derivative of ln[f(x)] = f'(x)/f(x). e.g. f(x) = x^2

ln(x^2) => 2x/x^2 and so on.

[hq319]

did you forget the 1 over root 2 ? lol

[kanopopo]

im scared of collage now

[markov2b1]

@goamira

first factor out the 25 from the inside the square root. this gives you 1/5 * integral ( 1 / sqrt (1- (x/5)^2)) , now u = x/5

so 5du = dx, so the 1/5 and 5 cancel. you are left with integral 1 / sqrt ( 1 - u^2) = arcsin u + c , or arcsin x/5 + c

[markov2b1]

very clever. great video!!! 

[goamira]

how do you solve the integral 1/(sqrt(25-x^2))dx

[ImmanualKant1989]

Kool thank you!

[ksonggg]

why can't the end just be ln of the entire bottom times its derivative? since it's one over the denominator.

[tamaratm22]

:D :D :D thanks!!!!!!

[sheazer]

his answer is wrong. in the second integration U must be 2X not X. check wolfram (the online integral solver, just google it) and the answer is different.