Asked by Monica
Evaluate the integral when x >0
∫ln(x^2+10x+16)dx
∫ln(x^2+10x+16)dx
Answers
Answered by
mathhelper
∫ln(x^2+10x+16)dx
= ∫ln(x^2 + 10x + 25 - 9) dx
= ∫ln( (x+5)^2 - 9) dx
= ∫ln ((x+5+3)(x+5-3))dx
= ∫ln( (x+8)(x+2) )dx
= ∫ ln( (x+8) + ln(x+2) ) dx
= ∫ln(x+8)dx + ∫ln(x+2) dx
= (x+8)ln(x+8) - x + (x+2)ln(x+2) - x + C
= (x+8)ln(x+8) + (x+2)ln(x+2) - 2x + C
= ∫ln(x^2 + 10x + 25 - 9) dx
= ∫ln( (x+5)^2 - 9) dx
= ∫ln ((x+5+3)(x+5-3))dx
= ∫ln( (x+8)(x+2) )dx
= ∫ ln( (x+8) + ln(x+2) ) dx
= ∫ln(x+8)dx + ∫ln(x+2) dx
= (x+8)ln(x+8) - x + (x+2)ln(x+2) - x + C
= (x+8)ln(x+8) + (x+2)ln(x+2) - 2x + C
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