Answer to Question #186230 in Calculus for Phyroe

Question

Integration by Partial Fractions

1.) ∫(8t^3 + 13)dt/(t+2)(4t^2+1)

2.) ∫(5x^2 -3x+ 18)dx/x(9-x^2) from 1 to 2

Answer 1

Question 1:

Given

$$\displaystyle\int \frac{8t^3+13}{(4t^2+1)(t+2)} dt$$

Since the degree of the numerator is not less than the degree of the denominator, we perform long polynomial division to yield:

$$\frac{8t^3+13}{(t+2)(4t^2+1)} = 2 + \frac{-16t^2-2t+9}{(4t^2+1)(t+2)}$$

Next, perform partial fraction decomposition on the fraction:

$$\frac{-16t^2-2t+9}{(4t^2+1)(t+2)}$$

Let

$$\frac{-16t^2-2t+9}{(4t^2+1)(t+2)}= \frac{At+B}{4t^2+1} + \frac{C}{t+2}$$

Upon simplifying,

$$-16t^2-2t+9 = (At+B)(t+2) + C(4t^2+1)$$

Expand the right-hand side

$$-16t^2-2t+9 = t^2(A+4C)+t(2A+B)+(2B+C)$$

On comparing the coefficients of $x^2$ , $x$ and the constants terms on both sides, we obtain the following system of equations:

$$-16 = A+4C; \quad -2 = 2A+B; \quad 9=2B+C$$

Solving the three equations simultaneously, we get $A = -4, B=6$ and $C=-3$

Therefore,

$$\frac{-16t^2-2t+9}{(4t^2+1)(t+2)}= 2 + \frac{6-4t}{4t^2+1} + \frac{-3}{t+2}$$

And

$$\displaystyle\int \frac{8t^3+13}{(4t^2+1)(t+2)} dt = \displaystyle\int \left(2 + \frac{6-4t}{4t^2+1} -\frac{3}{t+2} \right) dt$$

where,

$$\displaystyle\int 2\, dt = 2t+c_1 ; \displaystyle\int \left( -\frac{3}{t+2} \right) dt = -3\ln|t+2| + c_2$$

And

$$\displaystyle\int \left( \frac{6-4t}{4t^2+1} \right) dt = \displaystyle\int \frac{6}{4t^2+1} \, dt -\displaystyle\int\frac{4t}{4t^2+2} dt$$

Let $u=2t \Rightarrow \frac{du}{2}=dt.$ Similarly, let $v=4t^2+1 \Rightarrow \frac{dv}{2}=4t\,dt$

Therefore,

$$\displaystyle\int \frac{6}{4t^2+1} \, dt -\displaystyle\int\frac{4t}{4t^2+2} dt = \displaystyle\int \frac{6}{2(u^2+1)} \, du -\displaystyle\int\frac{1}{2}\frac{1}{v} \,dv$$

$$= 3\tan^{-1} (u) -\frac{1}{2} \ln|v| + c_3$$

$$= 3\tan^{-1} (2t) -\frac{1}{2} \ln|4t^2+1| + c_3$$

Upon combining the different integral results, we obtain

$$\displaystyle\int \frac{8t^3+13}{(4t^2+1)(t+2)} dt = 2t+3\tan^{-1}(2t) -\frac{1}{2}\ln|4t^2+1|-3\ln|t+2| + C$$

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Question 2:

Given

$$\displaystyle\int_1^2 \frac{5x^2-3x+18}{x(9-x^2)} \, dx$$

Factor the denominator and simplify the integrand as follows:

$$\frac{5x^2-3x+18}{x(9-x^2)} = \frac{-5x^2+3x-18}{x(9-x^2)} = \frac{-5x^2+3x-18}{x(x-3)(x+3)}$$

By partial fraction decomposition, let

$$\frac{-5x^2+3x-18}{x(x-3)(x+3)} = \frac{A}{x} + \frac{B}{x-3}+\frac{C}{x+3}$$

Simplify by multiplying both sides by $x(x-3)(x+3).$ This gives

$$-5x^2+3x-18 = x^2 (A+B+C) + 3(B-C)x - 9A$$

On comparing the coefficients of $x^2$ , $x$ and the constants terms on both sides, we obtain the following system of equations:

$$-5 = A+B+C ; \quad 3=3(B-C) ;\quad -18=-9A$$

Upon solving these three equations simultaneously, we get $A=2, B=-3$ and $C=-4$

Therefore,

$$\frac{-5x^2+3x-18}{x(x-3)(x+3)} = \frac{2}{x} + \frac{-3}{x-3}+\frac{-4}{x+3}$$

And

$$\displaystyle\int_1^2\frac{-5x^2+3x-18}{x(x-3)(x+3)} \, dx = \displaystyle\int_1^2 \left(\frac{2}{x} + \frac{-3}{x-3}+\frac{-4}{x+3}\right)\, dx$$

$$=\left[ 2\ln|x| -3\ln|x-3| - 4\ln|x+3| \right]^2_1$$

$$=\left[ 2\ln(2) -3\ln(1) - 4\ln(5) \right] - \left[ 2\ln(1) -3\ln(2) - 4\ln(4) \right]$$

$$= 5\ln(2) - 4\ln(5)+ 4\ln(4) = 2.5732$$