how would i integrate the following... int ( (1x+1)/(x^2+1) +( 1x+0)/(x^2+1)^2) ) ??
M missyc8 New member Joined Sep 7, 2009 Messages 13 Sep 23, 2009 #1 how would i integrate the following... int ( (1x+1)/(x^2+1) +( 1x+0)/(x^2+1)^2) ) ??
P PAULK Junior Member Joined Dec 13, 2007 Messages 124 Sep 23, 2009 #2 missyc8 said: how would i integrate the following... int ( (1x+1)/(x^2+1) +( 1x+0)/(x^2+1)^2) ) ?? Click to expand... Assuming this is what you mean: x+1 x -------- + ---------- (x^2+1) (x^2+1)^2 then write: x 1 x -------- + ------- + ---------- (x^2+1) (x^2+1) (x^2+1)^2 Integrate the first and third terms by u = x^2 + 1 and the second as an arctan(x).
missyc8 said: how would i integrate the following... int ( (1x+1)/(x^2+1) +( 1x+0)/(x^2+1)^2) ) ?? Click to expand... Assuming this is what you mean: x+1 x -------- + ---------- (x^2+1) (x^2+1)^2 then write: x 1 x -------- + ------- + ---------- (x^2+1) (x^2+1) (x^2+1)^2 Integrate the first and third terms by u = x^2 + 1 and the second as an arctan(x).
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Sep 24, 2009 #3 \(\displaystyle \int\bigg[\frac{x+1}{x^{2}+1}+\frac{x}{(x^{2}+1)^{2}}\bigg]dx\) \(\displaystyle = \ \int\frac{x}{x^{2}+1}dx+\int\frac{dx}{x^{2}+1}+\int\frac{x}{(x^{2}+1)^{2}}dx\) \(\displaystyle = \ \frac{ln(x^{2}+1)}{2}+arctan(x)-\frac{1}{2(x^{2}+1)}+C\)
\(\displaystyle \int\bigg[\frac{x+1}{x^{2}+1}+\frac{x}{(x^{2}+1)^{2}}\bigg]dx\) \(\displaystyle = \ \int\frac{x}{x^{2}+1}dx+\int\frac{dx}{x^{2}+1}+\int\frac{x}{(x^{2}+1)^{2}}dx\) \(\displaystyle = \ \frac{ln(x^{2}+1)}{2}+arctan(x)-\frac{1}{2(x^{2}+1)}+C\)
