Using Integration by Partial Fractions determine the integrals. 1)\ \int \frac...

Question:

Using Integration by Partial Fractions determine the integrals.

{eq}1)\ \int \frac {(15x+5)}{(x^2+5x)} \ dx\\ 2)\ \int\frac {(x^4- 2x^3+ 6x^2- 11x +2)}{(x^3- 3x^2+ 2x)} \ dx {/eq}

Integration by partial fraction:

The above question concerns the topic of integration by partial fraction. In this method, first, we check that numerator power is less than denominator power or not. If not, then we divide the expression and make the numerator power, less than the denominator. After that, we factorize the denominator in the liner factor or quadratic factor. The following are the different partial fraction forms.

{eq}\displaystyle \frac{px+q}{(x+a)(x+b)}=\frac{A}{x+a}+\frac{B}{x+b}\\ \displaystyle \frac{px+q}{(x+a)(x^2+bx+c)}=\frac{A}{x+a}+\frac{Bx+C}{x^2+bx+c}\\ {/eq}

To find the values of A, B, C, we compare coefficients of both side and then solve the equations.

Using Integration by Partial Fractions determine the integrals.

1.

{eq}\begin{align} \displaystyle &\int \frac {(15x+5)}{(x^2+5x)} \ dx\\ &\text{First we do the partial fraction of given function, we have} \\ \displaystyle \frac {(15x+5)}{x(x+5)} &= \frac{A}{x}+\frac{B}{x+5}\\ \displaystyle \frac {(15x+5)}{x(x+5)} &= \frac{A(x+5)+Bx}{x(x+5)}\\ &\text{Comparing both side ,we have}\\ 5A=5 &\rightarrow A=1\\ A+B=15 &\rightarrow B=14\\ \displaystyle \int \frac {(15x+5)}{(x^2+5x)} \ dx &= \int \frac{1}{x} dx + \int \frac{14}{x+5} dx \\ \displaystyle &= \ln |x| +14 \ln |x+5| + C \end{align} {/eq}

2.

{eq}\begin{align} \displaystyle &\int\frac {(x^4- 2x^3+ 6x^2- 11x +2)}{(x^3- 3x^2+ 2x)} \ dx \\ &\text{Dividing Numerator by denominator, we have}\\ \displaystyle &=\int x+1 +\frac {(7x^2- 13x +2)}{(x^3- 3x^2+ 2x)} \ dx---(1) \\ &\text{Now first, doing the partial fraction the last term, we have}\\ \displaystyle \frac {(7x^2- 13x +2)}{(x^3- 3x^2+ 2x)} &=\frac {(7x^2- 13x +2)}{(x(x^2- 3x+ 2)}\\ \displaystyle \frac {(7x^2- 13x +2)}{(x((x- 1)(x-2)} &= \frac{A}{x} +\frac{B}{x-1}+\frac{C}{x-2}---(2)\\ \displaystyle \frac {(7x^2- 13x +2)}{(x((x- 1)(x-2)} &= \frac{A(x-1)(x-2)+Bx(x-2)+Cx(x-1)}{x((x- 1)(x-2)}\\ \displaystyle \frac {(7x^2- 13x +2)}{(x((x- 1)(x-2)} &= \frac{Ax^2-3Ax+2A+Bx^2-2Bx+Cx^2-Cx}{x((x- 1)(x-2)}\\ \displaystyle \frac {(7x^2- 13x +2)}{(x((x- 1)(x-2)} &= \frac{(A+B+C)x^2-(3A+2B+C)x+2A}{x((x- 1)(x-2)}\\ &\text{Comparing the coefficients, we have}\\ 2A=2 \rightarrow &A=1 \\ 3A+2B+C=13 &\rightarrow 2B+C=10 \\ A+B+C=7 &\rightarrow B+C=6\\ &\text{Solving above equations we have}\\ B=4 \text{ and } & C=2\\ &\text{Substituting the values of A, B, C in equation (2), we have}\\ \displaystyle \frac {(7x^2- 13x +2)}{(x((x- 1)(x-2)} &= \frac{1}{x} +\frac{4}{x-1}+\frac{2}{x-2}\\ &\text{Now substituting above in equation (1) and then integrating, we have}\\ \displaystyle \int\frac {(x^4- 2x^3+ 6x^2- 11x +2)}{(x^3- 3x^2+ 2x)} \ dx &= \int \left ( x+1 + \frac{1}{x} +\frac{4}{x-1}+\frac{2}{x-2} \right )dx \\ \displaystyle &=\frac{x^2}{2} +x + \ln |x| +4 \ln |x-1| +2\ln |x-2| +C \end{align} {/eq}