Elvaluate the impropper integral or show that it diverges. Evaluate the integral if it converges `int_0^oo (1)/(x^2+2x+10)dx`



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Posted on (Answer #2)

`int_0^oo dx/(x^2+20+10)=int_0^oo dx/((x+1)^2+9)=`

`1/9int_0^oo dx/((x+1)^2/9+1)=1/9int_0^oo dx/(((x+1)/3)^2+1)=|(t=(x+1)/3),(dt=dx/3=>dx=3dt),(t_1=(0+1)/3=1/3),(t_2=oo)|`` `

`3/9int_(1/3)^oo dx/(t^2+1)=1/3arctan t|_(1/3) ^oo=`

`1/3(lim_(t->oo)arctan t-arctan(1/3)=1/3(pi/2-arctan(1/3))`

Your solution isĀ ` ` `1/3(pi/2-arctan(1/3)) approx 0.416349`

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