Friday

January 30, 2015

January 30, 2015

Posted by **Anonymous** on Tuesday, December 4, 2012 at 4:00am.

- Calculus -
**Steve**, Tuesday, December 4, 2012 at 5:00amlong division shows that we have

∫ 1 + (x-1)/(4x^2-4x+3)

= ∫ 1 + (x-1)/[(2x-1)^2 + 2] dx

let (2x-1)^2 = 2tan^2θ, so

(2x-1)^2 + 2 = 2tan^2θ+2 = 2sec^2θ

2x-1 = √2 tanθ

x = (√2 tanθ+1)/2

x-1 = (√2 tanθ-1)/2

dx = 1/2 √2 sec^2θ

∫ 1 + (x-1)/[(2x-1)^2 + 2] dx

= ∫ (1 + (√2 tanθ-1)/(4sec^2θ)) 1/√2 sec^2θ dθ

. . .

= 1/8(x + 1/8 log(4x^2-4x+3) - 1/(4√22) arctan (2x-1)/√2

**Answer this Question**

**Related Questions**

Math/Calculus - How would I solve the following integral with the substitution ...

Calculus - Find the volume of the solid whose base is the region in the xy-plane...

Calculus II - Integrate using integration by parts (integral) (5-x) e^3x u = 5-x...

Calculus II/III - A. Find the integral of the following function. Integral of (x...

Calculus - integral -oo, oo [(2x)/(x^2+1)^2] dx (a) state why the integral is ...

Calculus - Use the symmetry of the graphs of the sine and cosine functions as an...

calculus - 8). Part 1 of 2: In the solid the base is a circle x^2+y^2=16 and the...

Calculus 2 - The question is: Evaluate the improper integral for a>0. The ...

Calculus - evaluate the integral or state that it diverges. Check if I did it ...

Integral calculus - You have not proved to me that the Integral of lnsinx w.r.t....