- #1
LearninDaMath
- 295
- 0
[itex]\int\frac{2}{(x+3)\sqrt{x+10}}dx[/itex]
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First thing would be u-substitution, finding what I can replace in terms of u:
let [itex]u=\sqrt{x+10}[/itex]
[itex]\frac{du}{dx}=\frac{1}{2}(x+10)^{\frac{1}{2}-\frac{2}{2}}(x+10)'[/itex]
[itex]du=\frac{1}{2\sqrt{x+10}}dx[/itex] → [itex]dx=2\sqrt{x+10}du[/itex]
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Then replace those substitutions into the integral and simplify what I can:
[itex]\int\frac{(2)(2)\sqrt{x+10}du}{(x+3)\sqrt{x+10}}[/itex] → [itex]4\int\frac{du}{(x+3)}[/itex]
Then realize I still have an x term and figure out a way to arrange the previous substitutions to eliminate the remaining x term
[itex][u^{2}-10=x][/itex] → [itex][u^{2}-10+3=x+3][/itex] → [itex][u^{2}-7=x+3][/itex]
[itex]4\int\frac{1}{u^{2}-7}du[/itex] ...And then partial fractions from here
Is this a correct track so far?
_____________________________________
First thing would be u-substitution, finding what I can replace in terms of u:
let [itex]u=\sqrt{x+10}[/itex]
[itex]\frac{du}{dx}=\frac{1}{2}(x+10)^{\frac{1}{2}-\frac{2}{2}}(x+10)'[/itex]
[itex]du=\frac{1}{2\sqrt{x+10}}dx[/itex] → [itex]dx=2\sqrt{x+10}du[/itex]
______________________________________
Then replace those substitutions into the integral and simplify what I can:
[itex]\int\frac{(2)(2)\sqrt{x+10}du}{(x+3)\sqrt{x+10}}[/itex] → [itex]4\int\frac{du}{(x+3)}[/itex]
Then realize I still have an x term and figure out a way to arrange the previous substitutions to eliminate the remaining x term
[itex][u^{2}-10=x][/itex] → [itex][u^{2}-10+3=x+3][/itex] → [itex][u^{2}-7=x+3][/itex]
[itex]4\int\frac{1}{u^{2}-7}du[/itex] ...And then partial fractions from here
Is this a correct track so far?