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Integrate: a)\int \frac{4x^{2} + 31x - 24}{x^{3} + x^{2} - 12x} dx b)\int \frac{2x^{2} - 22x +...

Question:

Integrate:

a) {eq}\displaystyle \int \frac{4x^{2} + 31x - 24}{x^{3} + x^{2} - 12x} dx {/eq}

b) {eq}\displaystyle \int \frac{2x^{2} - 22x + 84}{(x - 5)^{2} (x + 1)} dx {/eq}

c) {eq}\displaystyle \int \frac{2x^{2} + 3x - 50}{x^{3} + 5x} dx {/eq}

Integration by partial fraction:

The above question concerns the topic of the integration by partial fraction methods. following are the different partial fraction forms.

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

To find the partial fraction coefficient {eq}(A,B,C..etc) {/eq}, we compare the coefficient of both sides and then solve the equation to determine the coefficients.

After partial fraction, the integral is converted into simple form, whose direct integration is possible.

Answer and Explanation:

a)

{eq}\begin{align} \displaystyle &\int \frac{4x^{2} + 31x - 24}{x^{3} + x^{2} - 12x} dx \\ \displaystyle &=\int \frac{4x^{2} + 31x - 24}{x(x^{2} + x - 12)} dx \\ \displaystyle &=\int \frac{4x^{2} + 31x - 24}{x(x^{2} +4 x-3x - 12)} dx \\ \displaystyle &=\int \frac{4x^{2} + 31x - 24}{x(x+4)(x-3)} dx \\ &\text{Doing partial fraction, we have }\\ \frac{4x^{2} + 31x - 24}{x(x+4)(x-3)} &=\frac{A}{x}+\frac{B}{x+4}+\frac{C}{x-3}---(1)\\ \frac{4x^{2} + 31x - 24}{x(x+4)(x-3)} &=\frac{A(x^2+x-12)+B(x^2-3x)+C(x^2+4x)}{x(x+4)(x-3)}\\ \frac{4x^{2} + 31x - 24}{x(x+4)(x-3)} &=\frac{x^2(A+B+C)+x(A-3B+4C)+(-12A)}{x(x+4)(x-3)}\\ &\text{Comparing coefficient, we have}\\ \displaystyle -12A &= -24 \Rightarrow A=2\\ \displaystyle A+B+C &=4 \Rightarrow B+C=2--(2)\\ \displaystyle A-3B+4C &=31 \Rightarrow -3B+4C=29--(3)\\ &\text{Solving equation(2) and (3), we have}\\ &B=-3,C=5 \\ &\text{ Substituting the value in equation (1), and then integrating, we have} \\ \int \frac{4x^{2} + 31x - 24}{x(x+4)(x-3)}dx &=\int \left (\frac{2}{x}-\frac{3}{x+4}+\frac{5}{x-3} \right )dx\\ \displaystyle &=2\log |x| -3 \log |x+4|+5 \log |x-3|+C\\ \end{align} {/eq}

b)

{eq}\begin{align} \displaystyle &\int \frac{2x^{2} - 22x + 84}{(x - 5)^{2} (x + 1)} dx \\ &\text{Doing partial fraction, we have}\\ \frac{2x^{2} - 22x + 84}{(x - 5)^{2} (x + 1)} &= \frac{A}{x-5}+\frac{B}{(x-5)^2}+\frac{C}{x+1}---(1)\\ \frac{2x^{2} - 22x + 84}{(x - 5)^{2} (x + 1)} &= \frac{A(x-5)(x+1)+B(x+1)+C(x-5)^2}{(x-5)^2(x+1)}\\ \frac{2x^{2} - 22x + 84}{(x - 5)^{2} (x + 1)} &= \frac{A(x^2-4x-5)+B(x+1)+C(x^2-10x+25)^2}{(x-5)^2(x+1)}\\ \frac{2x^{2} - 22x + 84}{(x - 5)^{2} (x + 1)} &= \frac{x^2(A+C)+x(-4A+B-10C)+(-5A+B+25C)}{(x-5)^2(x+1)}\\ &\text{Comparing coefficient both side}\\ A+C &=2 ---(2)\\ -4A+B-10C &=-22 ---(3)\\ -5A+B+25C &= 84---(4) \\ &\text{Solving equation (2),(3), (4) we have}\\ &A=-1,B=4,C=3\\ &\text{ Substituting the value in equation (1), and then integrating, we have} \\ \int \frac{2x^{2} - 22x + 84}{(x - 5)^{2} (x + 1)} dx &= \int \frac{-1}{x-5}+\frac{4}{(x-5)^2}+\frac{3}{x+1}\\ &= -\log |x-5| -\frac{4}{(x-5)} +3\log |x+1| +C \end{align} {/eq}

c)

{eq}\begin{align} \displaystyle &\int \frac{2x^{2} + 3x - 50}{x^{3} + 5x} dx \\ \displaystyle &=\int \frac{2x^{2} + 3x - 50}{x(x^{2} + 5)} dx \\ &\text{Doing partial fraction, we have}\\ \displaystyle \frac{2x^{2} + 3x - 50}{x(x^{2} + 5)} &= \frac{A}{x}+\frac{Bx+C}{x^2+5}---(1))\\ \displaystyle \frac{2x^{2} + 3x - 50}{x(x^{2} + 5)} &= \frac{Ax^2+5A+Bx^2+Cx}{x(x^2+5)}\\ &\text{Comparing coefficient, we have}\\ C &=3 \\ 5A =-50 &\Rightarrow A=-10\\ A+B =2 &\Rightarrow B=12 \\ &\text{Substituting the value, we have}\\ \displaystyle \int \frac{2x^{2} + 3x - 50}{x(x^{2} + 5)} dx &= \int \left (\frac{-10}{x}+\frac{12x+3}{x^2+5} \right ) dx\\ &= \int \left (\frac{-10}{x}+\frac{12x}{x^2+5}+\frac{3}{x^2+5} \right ) dx---(2)\\ &\text{Solving each integral separably, we have} \\ \int \frac{-10}{x} dx &=-10 \log |x|+C_1---(3) & \left [ \int \frac{dx}{x} =\log |x|+C\right ]\\ &12 \int \frac{x dx}{x^2+5} &\left [ \text{Solving integration by substitution} \right ]\\ \text{Let }x^2+5=t &\Rightarrow xdx=\frac{dt}{2} \text{ so}\\ 12\int \frac{\frac{dt}{2}}{t} &=6\log |t| &\left [ \text{ Substituting the value of t, we have} \right ]\\ 12 \int \frac{x dx}{x^2+5} &= 6\log |x^2+5|+C_2---(4)\\ \int \frac{3}{x^2+5} dx &= \frac{3}{\sqrt{5}}\tan^{-1} \frac{x}{\sqrt{5}}+C_3---(5) &\left [ \int \frac{dx}{x^2+a^2}= \frac{1}{a}\tan^{-1}\frac{x}{a}+C \right ]\\ &\text{So the solution of the given integral is }\\ \displaystyle \int \frac{2x^{2} + 3x - 50}{x^{3} + 5x} dx &= -10 \log |x|+6\log |x^2+5| + \frac{3}{\sqrt{5}}\tan^{-1} \frac{x}{\sqrt{5}} +C &\left [ C_1+C_2+C_3=C \right ]\\ \end{align} {/eq}


Learn more about this topic:

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How to Integrate Functions With Partial Fractions

from Math 104: Calculus

Chapter 13 / Lesson 9
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