Question #110639

Integrated ×^2+×+5/(×^2+4)(×+1) dx.

Integration by part

Expert's answer

Problem: x2+x+5(x2+4)(x+1)dx\int {\frac {x^2+x+5}{(x^2+4)(x+1)}dx}

Solution:

(1) x2+x+5(x2+4)(x+1)dx=x2+4+x+1(x2+4)(x+1)dx=I1+I2\int {\frac {x^2+x+5}{(x^2+4)(x+1)}dx}=\int {\frac {x^2+4+x+1}{(x^2+4)(x+1)}dx}=I_1+I_2

(2) I1=1x+1dx=ln(x+1)+CI_1=\int {\frac {1}{x+1}dx}=ln(x+1)+C

I2=1x2+4dxI_2=\int {\frac {1}{x^2+4}\cdot dx}

To evaluate I2I_2 one can use the substitution

(3) x=2tan(y)x=2tan(y) then

x2+4=4(tan2(y)+1)=4(sin2(y)+cos2(y)cos2(y))=4cos2(y);dx=2dycos2(y)x^2+4=4(tan^2(y)+1)=4(\frac{sin^2(y)+cos^2(y)}{cos^2(y)})=\frac{4}{cos^2(y)};\\ dx=\frac{2dy}{cos^2(y)} and

(4) I2=14/cos2(y)2dycos2(y)=12dy=y2+CI_2=\int{\frac{1}{4/cos^2(y)}\cdot \frac{2dy}{cos^2(y)}}=\frac{1}{2}\int{dy}=\frac{y}{2}+C

Reverse substitution (3) we have y=arctan(x2)y=arctan(\frac{x}{2}).

In notation  John Herschel we may write y=tan1(x2)y= tan^{-1}(\frac{x}{2}).

(5) I2=12arctan(x2)+CI_2=\frac{1}{2}arctan(\frac{x}{2})+C

Substitute (2) and (5) in (1) we get the solution to the problem

Answer:x2+x+5(x2+4)(x+1)dx=ln(x+1)+12arctan(x2)+C\int {\frac {x^2+x+5}{(x^2+4)(x+1)}dx}=ln(x+1)+\frac{1}{2}arctan(\frac{x}{2})+C

Comment:

(1) The solution method used here is not called "integration by parts", although the integral is divided into two parts for integration. The correct name for this method is 'fractional part integration" or 'Integration using Partial Fractions'. This method used for rational function integrals http://www.math.wsu.edu/faculty/genz/140/lessons/l506.pdf.

(2) If we use y=arctan(x2)y=arctan(\frac{x}{2}) in (3) we can immediately get

dy=arctanx(x2)dx=dx211+(x/2)2=2dx4+x2dy=arctan'_x(\frac{x}{2})dx=\frac{dx}{2}\cdot \frac{1}{1+(x/2)^2}=\frac{2dx}{4+x^2} and

dx4+x2=dy2\frac{dx}{4+x^2}=\frac{dy}{2}



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