Evaluate the following integral ∫(3x-4)/√(2x^2-7x+5) dx
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Mathematics
3x-4=λ(4x-7)+μ
need an explanation for this answer? contact us directly to get an explanation for this answer3=4λ ,-4=-7λ+µ then λ=3/4 and μ=5/4
I=∫(3/4 (4x-7)+5/4)/√(2x^2-7x+5) dx=3/4 ∫(4x-7)/√(2x^2-7x+5) dx+5/4 ∫dx/√(2x^2-7x+5)
=3/4 ∫(4x-7) (2x^2-7x+5)^(-1/2) dx+5/(4√2) ∫dx/√((x-7/4)^2-9/16) dx
=3/4 √(2x^2-7x+5)+5/(4√2) cosh^(-1)〖((4x-7))/3〗+c