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Question
Evaluate the following.
`int x/(4x^4 - 20x^2 - 3)dx`
Solution
I = `intx/(4x^4 - 20x^2 - 3)dx`
= `intx/(4[(x^2)^2 - 5x^2 - 3/4])dx`
= `1/4intx/((x^2)^2 - 5x^2 - 3/4)dx`
Put x2 = t
diff. w.r.t x both sides,
2x = `(dt)/(dx)`
2x.dx = dt
`x.dx = 1/2 dt`
I = `1/4int(1/2.dt)/(t^2 - 5t - 3/4)`
I = `1/4. 1/2int 1/(t^2 - 5t - 3/4)dt`
add and subtract `(1/2 xx -5)^2 = 25/4`
I = `1/8int1/((t^2 - 5t + 25/4) - 3/4 - 25/4)dt`
= `1/8int1/((t - 5/2)^2 - 7)dt`
= `1/8int1/((t - 5/2)^2 - (sqrt7)^2)dt`
I = `1/8. 1/(2(sqrt7)). log(|(t - 5/2 - sqrt7)/(t - 5/2 + sqrt7)|) + c`
= `1/(16sqrt7) . log|((2t)/2 - 5/2 - (2sqrt7)/2)/((2t)/2 - 5/2 + (2sqrt7)/2)| + c`
= `1/(16sqrt7) . log|(2t - 5 - 2sqrt7)/(2t - 5 + 2sqrt7)| + c`
I = `1/(16sqrt7) . log|(2x^2 - 5 - 2sqrt7)/(2x^2 - 5 + 2sqrt7)| + c`
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