- #1
DDarthVader
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Hello! I'm doing some calculus exercises and I'm having some difficulty to find the right answer.
The integral is [itex]\int \frac{1}{\sqrt{5x+8}}dx[/itex] and the exercise tell me to solve this by using (a) [itex]u=5x+8[/itex] and (b) [itex]u=\sqrt{5x+8}[/itex]
This is my attempt:
(a) Well, [itex]u=5x+8[/itex] then [itex]\frac{du}{dx}=5[/itex] which means [itex]dx=\frac{du}{5}[/itex]. Then my integral is now [itex]\int \frac{du}{5 \sqrt{u}} = \frac{1}{5}ln_{\sqrt{u}} = \frac{1}{5}ln_{\sqrt{5x+8}} = \frac{1}{5}ln_{5x}ln_{8}[/itex] Is this correct?
(b) Now [itex]u=\sqrt{5x+8}[/itex]. Then [itex]\frac{du}{dx}=\frac{1}{2\sqrt{5x+8}}[/itex] which means [itex]dx=2u[/itex] and now my integral is [itex]2 \int du = u^2 = 5x+8[/itex]. Is this correct?
Thanks!
Homework Statement
The integral is [itex]\int \frac{1}{\sqrt{5x+8}}dx[/itex] and the exercise tell me to solve this by using (a) [itex]u=5x+8[/itex] and (b) [itex]u=\sqrt{5x+8}[/itex]
Homework Equations
The Attempt at a Solution
This is my attempt:
(a) Well, [itex]u=5x+8[/itex] then [itex]\frac{du}{dx}=5[/itex] which means [itex]dx=\frac{du}{5}[/itex]. Then my integral is now [itex]\int \frac{du}{5 \sqrt{u}} = \frac{1}{5}ln_{\sqrt{u}} = \frac{1}{5}ln_{\sqrt{5x+8}} = \frac{1}{5}ln_{5x}ln_{8}[/itex] Is this correct?
(b) Now [itex]u=\sqrt{5x+8}[/itex]. Then [itex]\frac{du}{dx}=\frac{1}{2\sqrt{5x+8}}[/itex] which means [itex]dx=2u[/itex] and now my integral is [itex]2 \int du = u^2 = 5x+8[/itex]. Is this correct?
Thanks!