∫√(4x2+40x)dx
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∫√(4x²+40x)dx
= 2∫√(x²+10x)dx
= 2∫√((x+5)²-25)dx
= 2∫√(u²-25)du , u=x+5
This is a standard integral, which you can look up in a table. You find
2∫√(u²-25)du = u √(u²-25) - 25 ln (u+√(u²-25)) +c, so that
∫√(4x²+40x)dx= (x+5) √((x+5)²-25) - 25 ln (x+5+√((x+5)²-25)) +c
= (x+5) √(x²+10x) - 25 ln (x+5+√(x²+10x)) +c
Let me know if you want to see how the formula for the standard integral can be derived.