# Evaluate the indefinite integral. \int \frac{25x^3}{(x - 1)^2(x^2 + 4)} dx (a) 8\ln |x - 1| +...

## Question:

Evaluate the indefinite integral.

{eq}\displaystyle\int \frac{25x^3}{\left(x - 1\right)^2\left(x^2 + 4\right)}\,dx {/eq}

(a) {eq}8\ln \left|x - 1\right| + 6\ln \left(x^2 + 4\right) + 16\tan^{-1}\left(\frac{x}{2}\right) + C {/eq}

(b) {eq}5\ln \left|x - 1\right| - \frac{5}{x - 1} + 6\ln \left(x^2 + 4\right) + 4\tan^{-1}\left(\frac{x}{2}\right) + C {/eq}

(c) {eq}13\ln \left|x - 1\right| - \frac{5}{x - 1} + 16\tan^{-1}\left(\frac{x}{2}\right) + C {/eq}

(d) {eq}13\ln \left|x - 1\right| - \frac{5}{x - 1} + 6\ln \left(x^2 + 4\right) + 16\tan^{-1}\left(\frac{x}{2}\right) + C {/eq}

## Answer and Explanation:

We need to find the value of {eq}I = \int {\frac{{25{x^3}}}{{{{\left( {x - 1} \right)}^2}\left( {{x^2} + 4} \right)}}dx} {/eq}

Since this is a rational fraction and the roots of the denominator {eq}Q \left( x \right) {/eq} are known, therefore we can apply partial fractions here.

Also, some of the roots of the denominator are repeated and some of them are non-linear, therefore the partial fraction of the above rational function is given by

{eq}\dfrac{{25{x^3}}}{{{{\left( {x - 1} \right)}^2}\left( {{x^2} + 4} \right)}} = \dfrac{A}{{x - 1}} + \dfrac{B}{{{{\left( {x - 1} \right)}^2}}} + \dfrac{{Cx + D}}{{{x^2} + 4}} {/eq}.

This is true for all the values of {eq}x {/eq}.

Let us find the values of {eq}A,B, C \text{ and } D {/eq} using different values of {eq}x. {/eq}

{eq}\eqalign{ & \Rightarrow \dfrac{{25{x^3}}}{{{{\left( {x - 1} \right)}^2}\left( {{x^2} + 4} \right)}} = \dfrac{{A\left( {x - 1} \right)\left( {{x^2} + 4} \right) + B\left( {{x^2} + 4} \right) + \left( {Cx + D} \right){{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}\left( {{x^2} + 4} \right)}} \cr & \Rightarrow 25{x^3} = A\left( {x - 1} \right)\left( {{x^2} + 4} \right) + B\left( {{x^2} + 4} \right) + \left( {Cx + D} \right){\left( {x - 1} \right)^2} \cr} {/eq}

Now,

{eq}\eqalign{ & {\text{For }}x = 1 \cr & 25 = B\left( 5 \right) \cr & B = 5 \cr & {\text{For }}x = 0 \cr & 0 = A\left( { - 1} \right)\left( 4 \right) + B\left( 4 \right) + \left( D \right)\left( 1 \right) \cr & 0 = - 4A + 20 + D \cr & 4A - D = 20......................\left( 1 \right) \cr & {\text{For }}x = 2 \cr & 200 = A\left( 8 \right) + 5\left( 8 \right) + \left( {2C + D} \right) \cr & 160 = 8A + 2C + D............(2) \cr & {\text{For }}x = - 1 \cr & - 25 = - 10A + 5\left( 5 \right) + \left( { - C + D} \right)\left( 4 \right) \cr & - 25 = - 10A + 25 - 4C + 4D \cr & - 50 = - 10A - 4C + 4D..........\left( 3 \right) \cr} {/eq}

Solving {eq}1, 2 \text{ and } 3 {/eq}, we get {eq}A=13, C= 12 \text{ and } D=32 {/eq}

Therefore, the partial fraction is given by

{eq}\dfrac{{25{x^3}}}{{{{\left( {x - 1} \right)}^2}\left( {{x^2} + 4} \right)}} = \dfrac{{13}}{{x - 1}} + \dfrac{5}{{{{\left( {x - 1} \right)}^2}}} + \dfrac{{12x + 32}}{{{x^2} + 4}} {/eq}

Integrating, we get

{eq}\eqalign{ & \int {\dfrac{{25{x^3}dx}}{{{{\left( {x - 1} \right)}^2}\left( {{x^2} + 4} \right)}}} = \int {\dfrac{{13}}{{x - 1}}dx} + \int {\dfrac{5}{{{{\left( {x - 1} \right)}^2}}}} + \int {\left( {\dfrac{{12x + 32}}{{{x^2} + 4}}} \right)} dx \cr & \int {\dfrac{{25{x^3}dx}}{{{{\left( {x - 1} \right)}^2}\left( {{x^2} + 4} \right)}}} = 13\ln \left| {x - 1} \right| - \dfrac{5}{{\left( {x - 1} \right)}} + 6\int {\dfrac{{2x}}{{{x^2} + 4}}} dx + 32\int {\dfrac{{dx}}{{{x^2} + 4}}} \cr & \int {\dfrac{{25{x^3}dx}}{{{{\left( {x - 1} \right)}^2}\left( {{x^2} + 4} \right)}}} = 13\ln \left| {x - 1} \right| - \dfrac{5}{{\left( {x - 1} \right)}} + 6\ln \left| {{x^2} + 4} \right| + \frac{{32}}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) \cr & \int {\dfrac{{25{x^3}dx}}{{{{\left( {x - 1} \right)}^2}\left( {{x^2} + 4} \right)}}} = \color{red}{13\ln \left| {x - 1} \right| - \dfrac{5}{{\left( {x - 1} \right)}} + 6\ln \left| {{x^2} + 4} \right| + 16{\tan ^{ - 1}}\left( {\dfrac{x}{2}} \right) + C} \cr} {/eq}

Hence, {eq}\color{red}{d} {/eq} is the correct option.