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Methods of integration (view source)
Revision as of 18:06, December 19, 2018
, 18:06, December 19, 2018→Algebraic Substitution: Tidy up formatting
Given an integral
Given an integral
:<big><math>\int\frac{2x}{x^2+3}dx</math></big>
:<math>\int\frac{2x}{x^2+3}dx</math>
We can substitute the term :<big><math>x^2+3</math></big> with a u. Giving us
We can substitute the term ''x''<sup>2</sup>+3 with a ''u''. Giving us
:<big><math>u=x^2+3</math></big>
:<math>u=x^2+3</math>
We then take the derivative of u with respect to x,
We then take the derivative of ''u'' with respect to ''x'',
:<big><math>\frac{du}{dx}=2x</math></big>
:<math>\frac{du}{dx}=2x</math>
We then set the terms equal to du,
We then set the terms equal to d''u'',
:<big><math>du=2xdx</math></big>
:<math>du=2xdx</math>
Now we are ready to rewrite the integral,
Now we are ready to rewrite the integral,
:<big><math>\int\frac{2x}{x^2+3}dx=\int\frac{1}{u}du</math></big>
:<math>\int\frac{2x}{x^2+3}dx=\int\frac{1}{u}du</math>
We can rewrite the integral this way due to the substitution of the x terms with the u terms.
We can rewrite the integral this way due to the substitution of the ''x'' terms with the ''u'' terms.
Now we can solve the integral in terms of u.
Now we can solve the integral in terms of ''u''.
:<big><math>\int\frac{1}{u}du=ln|u|+c</math></big>
:<math>\int\frac{1}{u}du=ln|u|+c</math>
Now we replace u with the term :<big><math>x^2+3</math></big> to get,
Now we replace ''u'' with the term ''x''<sup>2</sup>+3 to get,
:<big><math>ln|x^2+3|+c</math></big>
:<math>\ln|x^2+3|+c</math>
We can check this by taking the derivative of :<big><math>ln|x^2+3|</math></big>,
We can check this by taking the derivative of <math display="inline">\ln|x^2+3|</math>,
:<big><math>\frac{d}{dx}ln|x^2+3|=\left (\frac{1}{x^2+3} \right)(2x)=\frac{2x}{x^2+3}</math></big>
:<math>\frac{d}{dx}\ln|x^2+3|=\left (\frac{1}{x^2+3} \right)(2x)=\frac{2x}{x^2+3}</math>
===Reverse Chain Rule===
===Reverse Chain Rule===