# Which of the following integrals cannot be integrated using partial fractions using linear...

## Question:

Which of the following integrals cannot be integrated using partial fractions using linear factors with real coefficients?

{eq}a) \int \frac{(x^2-1)}{(x^3+x)} dx \\b) \int \frac{1}{(9x^2-4)} dx \\c) \int \frac{(x^3-x+3)}{(x^2+x-2)} dx {/eq} d)All of these

## Integration by partial fraction:

The above question concerns the topic of the integration by partial fraction. Followings are the various partial fraction form:

{eq}\displaystyle \frac{1}{(x+a)(x+b)}=\frac{A}{(x+a)}+\frac{B}{x+b}\\ \displaystyle \frac{1}{(x+a)(x^2+bx+c)}=\frac{A}{(x+a)}+\frac{Bx+C}{x^2+bx+c}\\ {/eq}

After doing partial fraction, given integral is converted into simpler form and can be evaluated directly or by using integration by substitution.

## Answer and Explanation:

Doing partial fraction of each part, we have

a)

{eq}\begin{align} \displaystyle I&=\int \frac{(x^2-1)}{(x^3+x)} dx \\ \displaystyle I&=\int \left [\frac{x^2}{(x^3+x)} - \frac{1}{ (x^3+x)} \right ] dx \\ \displaystyle I& =\int \left [\frac{x}{(x^2+1)} - \frac{1}{x (x^2+1)} \right ] dx---(1) \\ &\text{Doing partial fraction of second part we have}\\ \displaystyle \frac{1}{x (x^2+1)} &= \frac{A}{x}+\frac{Bx+C}{x^2+1}\\ \displaystyle \frac{1}{x (x^2+1)} &= \frac{A(x^2+1)+(Bx+C)x}{x(x^2+1)}\\ \displaystyle 1 &=(A+B)x^2+Cx+A\\ &\text{Comparing both side we have}\\ A &=1 ,\\ C &=0,\\ A+B=0 &\rightarrow B=-A\\ B &=-1\\ &\text{Substituting the values in equation (1), we have}\\ \displaystyle I& =\int \left [\frac{x}{(x^2+1)} - \frac{1}{x} +\frac{x}{x^2+1}\right ]dx\\ \displaystyle I& =\int \left [\frac{2x}{(x^2+1)} - \frac{1}{x} \right ]dx\\ \end{align} {/eq}

So this function can be integrated using partial fractions using linear factors with real coefficients.

b)

{eq}\begin{align} \displaystyle I&=\int \frac{1}{(9x^2-4)} dx \\ \displaystyle I&=\int \frac{1}{(3x-2)(3x+2)} dx---(1) \\ &\text{Doing partial fraction of the function we have}\\ \displaystyle \frac{1}{(3x-2)(3x+2)} &=\frac{A}{3x-2}+\frac{B}{3x+2}\\ \displaystyle \frac{1}{(3x-2)(3x+2)} &=\frac{A(3x+2)+B(3x-2)}{(3x-2)(3x+2)}\\ \displaystyle 1&=x(3A+3B)+2A-2B\\ &\text{Comparing both side we have}\\ 2A-2B &=1---(2) \text{ and}\\ 3A+3B&=0---(3)\\ &\text{Solving equation (2) and (3) we have,}\\ \displaystyle A &=\frac{1}{4}\\ \displaystyle B &=\frac{-1}{4}\\ &\text{Substituting values in equation(1), we have}\\ \displaystyle I&=\int \left [ \frac{1}{4(3x-2)}-\frac{1}{4(3x+2)} \right ] \end{align} {/eq}

So this function can be integrated using partial fractions using linear factors with real coefficients.

c)

{eq}\begin{align} \displaystyle I &=\int \frac{(x^3-x+3)}{(x^2+x-2)} dx \\ &\text{Dividing the function we have}\\ \displaystyle I &=\int \frac{(x^3-x+3)}{(x^2+x-2)} dx \\ \displaystyle I &=\int \left [(2x-1) + \frac{(2x+1)}{(x^2+x-2)} \right ] dx \\ \displaystyle I &=\int \left [(2x-1) + \frac{(2x+1)}{(x^2+2x-x-2)} \right ] dx \\ \displaystyle I &=\int \left [(2x-1) + \frac{(2x+1)}{((x+2)(x-1)} \right ] dx ---(1)\\ &\text{Doing partial fraction for the last part}\\ \displaystyle \frac{(2x+1)}{(x+2)(x-1)} &=\frac{A(x-1)+B(x+2)}{(x+2)(x-1)}\\ 2x+1&=x(A+B)+(-A+2B)\\ &\text{Comparing both side , we have}\\ A+B &=2---(2)\\ -A+2B &=1---(3)\\ &\text{Solving these two , we have} A=1 \text{ and } B=1\\ &\text{Substituting in equation(1), we have}\\ \displaystyle I &=\int \left [(2x-1) + \frac{1}{x+2} +\frac{1}{x-1} \right ] dx \end{align} {/eq}

So this function can be integrated using partial fractions using linear factors with real coefficients.

Clearly from all the parts, all of these can be integrated using partials fractions with linear factors and real coefficients