Question

If $\int{\dfrac{3{{z}^{3}}-8z+5}{\sqrt{{{z}^{2}}-4z-7}}}dz=\left( {{z}^{2}}+az+36 \right)\sqrt{{{z}^{2}}-4z-7}+b\log \left| z-2+\sqrt{{{z}^{2}}-4z-7} \right|+C$, where $a,b\in I$ and $C$ is the integration constant then

A. $a > b$

B. $a < b$

C. $a+b=117$

D. Exactly one out of $a$ or $b$ is a prime number.

Answer 1

Hint: We first take the L.H.S part and do the integration. While integrating we will use two substitutions, one is $z-2=u$ and $u=\sqrt{11}\sec v$ . After substituting each value, we have an equation consisting of multiple terms in integration. We will find the integration of each term individually using some integration formulas and substituting necessary values where they are required. After doing integration for all the individual terms we will add them to get the final integral value of given expression $\int{\dfrac{3{{z}^{3}}-8z+5}{\sqrt{{{z}^{2}}-4z-7}}dz}$ and then we will compare our result of integration with the R.H.S, then we will get the values of $a,b$.

Complete step by step answer:

Given that,

$\int{\dfrac{3{{z}^{3}}-8z+5}{\sqrt{{{z}^{2}}-4z-7}}}dz=\left( {{z}^{2}}+az+36 \right)\sqrt{{{z}^{2}}-4z-7}+b\log \left| z-2+\sqrt{{{z}^{2}}-4z-7} \right|+C$

Take L.H.S from the above given equation, then

$L.H.S=\int{\dfrac{3{{z}^{3}}-8z+5}{\sqrt{{{z}^{2}}-4z-7}}dz}.....\left( \text{i} \right)$

Simplify the expression ${{z}^{2}}-4z-7$ which is in denominator, then

$  {{z}^{2}}-4z-7={{z}^{2}}-2.2z+{{2}^{2}}-{{2}^{2}}-7 $

$  ={{z}^{2}}-2\left( 2 \right)\left( z \right)+{{2}^{2}}-4-7 $

$  ={{\left( z-2 \right)}^{2}}-11$

Let us take the substitution of ${{z}^{2}}-4z-7={{\left( z-2 \right)}^{2}}-11$ and $z-2=u$ then $dz=du$ in equation $\left( \text{i} \right)$

$\therefore  \int{\dfrac{3{{z}^{3}}-8z+5}{\sqrt{{{z}^{2}}-4z-7}}dz}=\int{\dfrac{3{{\left( u+2 \right)}^{3}}-8\left( u+2 \right)+5}{\sqrt{{{u}^{2}}-11}}}du $

$ =\int{\dfrac{3{{\left( u+2 \right)}^{3}}-8u-16+5}{\sqrt{{{u}^{2}}-11}}du} $

$ =\int{\dfrac{3{{\left( u+2 \right)}^{3}}-8u-11}{\sqrt{{{u}^{2}}-11}}du}$

Now take the substitution $u=\sqrt{11}\sec v$ then $v={{\sec }^{-1}}\left( \dfrac{u}{\sqrt{11}} \right)$ and $du=\sqrt{11}\sec v\tan v.dv$in the above equation, then we have

$   \int{\dfrac{3{{\left( u+2 \right)}^{3}}-8u-11}{\sqrt{{{u}^{2}}-11}}}du=\int{\dfrac{3{{\left( \sqrt{11}\sec v+2 \right)}^{3}}-8\left( \sqrt{11}\sec v \right)-11}{\sqrt{{{\left( \sqrt{11}\sec \right)}^{2}}-11}}}.\left( \sqrt{11}\sec v\tan v.dv \right) $

$ =\int{\dfrac{3{{\left( \sqrt{11}\sec v+2 \right)}^{3}}-8\sqrt{11}\sec v-11}{\sqrt{11\left( {{\sec }^{2}}-1 \right)}}}.\left( \sqrt{11}\sec v\tan v.dv \right) $

$ =\int{\dfrac{3{{\left( \sqrt{11}\sec v+2 \right)}^{3}}-8\sqrt{11}\sec v-11}{\sqrt{11}\tan v}.}\left( \sqrt{11}\sec v\tan v.dv \right)\text{ }\left[ \because {{\sec }^{2}}x-{{\tan }^{2}}x=1 \right] $

$  =\int{\left( 3\sec v{{\left( \sqrt{11}\sec v+2 \right)}^{3}}-8\sqrt{11}{{\sec }^{2}}v-11\sec v \right)dv} $

$  =3\int{\sec v}{{\left( \sqrt{11}\sec v+2 \right)}^{3}}.dv-8\sqrt{11}\int{{{\sec }^{2}}v}.dv-11\int{\sec v.dv}....\left( \text{ii} \right) $

Finding the integrals of $\sec v{{\left( \sqrt{11}\sec +2 \right)}^{3}}$ , ${{\sec }^{2}}v$ individually, then

$ {{\int{\sec v\left( \sqrt{11}\sec v+2 \right)}}^{3}}dv=\int{\sec v\left[ {{\left( \sqrt{11}\sec v \right)}^{3}}+{{2}^{3}}+3\left( \sqrt{11}\sec v \right)\left( 2 \right)\left( \sqrt{11}\sec v+2 \right) \right]}dv $

$  =\int{\sec v\left[ {{11}^{\dfrac{3}{2}}}{{\sec }^{3}}v+8+6\sqrt{11}\sec v\left( \sqrt{11}\sec v+2 \right) \right]} $

$  =\int{\sec v\left[ {{11}^{\dfrac{3}{2}}}{{\sec }^{3}}v+8+6.11{{\sec }^{2}}v+12\sqrt{11}\sec v \right]dv} $

$  =\int{\sec v\left[ {{11}^{\dfrac{3}{2}}}{{\sec }^{3}}v+66{{\sec }^{2}}v+12\sqrt{11}\sec v+8 \right]}dv $

$ =\int{\left( {{11}^{\dfrac{3}{2}}}{{\sec }^{4}}v+66{{\sec }^{3}}v+12\sqrt{11}{{\sec }^{2}}v+8\sec v \right)dv}....\left( \text{iii} \right) $

Now the values of $\int{{{\sec }^{4}}v}dv,\int{{{\sec }^{3}}v}dv,\int{{{\sec }^{2}}}dv$ are

For finding the value of $\int{{{\sec }^{2}}v}dv$ consider the equation $\dfrac{d}{dx}\left( \tan x+C \right)$, we have

$   \dfrac{d}{dx}\left( \tan x+C \right)=\dfrac{d}{dx}\left( \tan x \right)+\dfrac{d}{dx}\left( C \right) $

$  \dfrac{d}{dx}\left( \tan x+C \right)={{\sec }^{2}}x+0 $

$  d\left( \tan x+C \right)={{\sec }^{2}}xdx $

$  \int{{{\sec }^{2}}x}dx=\tan x+C.....\left( \text{a} \right) $

For finding the value of $\int{{{\sec }^{3}}x}dx$ use the integral by part then $p=\sec x,dq={{\sec }^{2}}x,dp=\sec x\tan xdx,q=\tan x$

Substitute the above values in $pq-\int{qdp}$, then

$  \int{{{\sec }^{3}}x}dx=\sec x\tan x-\int{{{\tan }^{2}}x\sec x}dx $

$  =\sec x\tan x-\int{\left( {{\sec }^{2}}x-1 \right)\sec xdx} $

$  =\sec x\tan x-\int{{{\sec }^{3}}x}dx+\int{\sec xdx} $

$ \int{{{\sec }^{3}}x}dx+\int{{{\sec }^{3}}x}dx=\sec x\tan x+\int{\sec xdx} $

$ 2\int{{{\sec }^{3}}x}dx=\sec x\tan x+\log \left| \sec x+\tan x \right| $

$ \int{{{\sec }^{3}}x}dx=\dfrac{1}{2}\sec x\tan x+\dfrac{1}{2}\log \left| \sec x+\tan x \right|+C....\left( \text{b} \right)$

Finding the value of $\int{{{\sec }^{4}}x}dx$.

$  \int{{{\sec }^{4}}x}dx=\int{{{\sec }^{2}}x}\left( {{\sec }^{2}}x \right)dx $

$  =\int{{{\sec }^{2}}x}\left( 1+{{\tan }^{2}}x \right)dx $

Let us substitute $t=\tan x$ and $dt={{\sec }^{2}}xdx$ in above equation, then

$ \int{{{\sec }^{4}}x}dx=\int{\left( 1+{{t}^{2}} \right)dt} $

$  =t+\dfrac{{{t}^{3}}}{3}+C $

$ =\tan x+\dfrac{{{\tan }^{3}}x}{3}+C....\left( \text{c} \right) $

From equations $a,b,c$ we have values of $\int{{{\sec }^{4}}v}dv,\int{{{\sec }^{3}}v}dv,\int{{{\sec }^{2}}}dv$, so substituting those values in equation $\left( \text{iii} \right)$ then we have

$ \int{\sec v{{\left( \sqrt{11}\sec v+2 \right)}^{3}}dv}={{11}^{\dfrac{3}{2}}}\int{{{\sec }^{4}}vdv}+66\int{{{\sec }^{3}}v}dv+12\sqrt{11}\int{{{\sec }^{2}}vdv}+8\int{\sec vdv} $

$  ={{11}^{\dfrac{3}{2}}}\left[ \tan v+\dfrac{{{\tan }^{3}}v}{3} \right]+66\left[ \dfrac{1}{2}\left( \sec v\tan v+\log \left| \sec v+\tan v \right| \right) \right]+12\sqrt{11}\left( \tan v \right)+8\log \left| \sec v+\tan v \right|+C $

$ ={{11}^{\dfrac{3}{2}}}\tan v+\dfrac{{{11}^{\dfrac{3}{2}}}}{3}{{\tan }^{3}}v+33\sec v\tan v+33\log \left| \sec v+\tan v \right|+12\sqrt{11}\tan v+8\log \left| \sec v+\tan v \right|+C $

$ =\dfrac{{{11}^{\dfrac{3}{2}}}}{3}{{\tan }^{3}}v+12\sqrt{11}\tan v+{{\left( \sqrt{11} \right)}^{3}}\tan v+33\sec v.\tan v+41\log \left| \sec v+\tan v \right|+C $

$ =\dfrac{{{11}^{\dfrac{3}{2}}}}{3}{{\tan }^{3}}v+\sqrt{11}\left[ 12+{{\left( \sqrt{11} \right)}^{2}} \right]\tan v+33\sec v.\tan v+41\log \left| \sec v+\tan v \right|+C $

$ =\dfrac{{{11}^{\dfrac{3}{2}}}}{3}{{\tan }^{3}}v+23\sqrt{11}\tan v+33\sec v.\tan v+41\log \left| \sec v+\tan v \right|+C$

Substituting the above value in equation $\left( \text{ii} \right)$, then we have

$ \int{\dfrac{3{{\left( u+2 \right)}^{3}}-8u-11}{\sqrt{{{u}^{2}}-11}}}=3\int{\sec v{{\left( \sqrt{11}\sec v+2 \right)}^{3}}dv}-8\sqrt{11}\int{{{\sec }^{2}}vdv}-11\int{\sec vdv} $

$ =3\left[ \dfrac{{{11}^{\dfrac{3}{2}}}}{3}{{\tan }^{3}}v+23\sqrt{11}\tan v+33\sec v.\tan v+41\log \left| \sec v+\tan v \right| \right]-8\sqrt{11}\tan v-11\log \left| \sec v+\tan v \right|+C $

$ ={{11}^{\dfrac{3}{2}}}{{\tan }^{3}}v+69\sqrt{11}\tan v+99\sec v.\tan v+123\log \left| \sec v+\tan v \right|-8\sqrt{11}\tan v-11\log \left| \sec v+\tan v \right|+C $

$ ={{11}^{\dfrac{3}{2}}}{{\tan }^{3}}v+61\sqrt{11}\tan v+99\sec v.\tan v+112\log \left| \sec v+\tan v \right|+C $

But we have the value of $v$ as ${{\sec }^{-1}}\left( \dfrac{u}{\sqrt{11}} \right)$ i.e. $\sec v=\dfrac{u}{\sqrt{11}}$ hence $\tan v=\sqrt{{{\sec }^{2}}v-1}=\sqrt{\dfrac{{{u}^{2}}}{11}-1}$, no the above equation is modified as

$\int{\dfrac{3{{\left( u+2 \right)}^{3}}-8u-11}{\sqrt{{{u}^{2}}-11}}}={{11}^{\dfrac{3}{2}}}\left( \sqrt{\dfrac{{{u}^{2}}}{11}-1} \right)+61\sqrt{11}\left( \sqrt{\dfrac{{{u}^{2}}}{11}-1} \right)+99\left( \dfrac{u}{\sqrt{11}} \right)\left( \sqrt{\dfrac{{{u}^{2}}}{11}-1} \right)+112\log \left| \left( \dfrac{u}{\sqrt{11}} \right)+\left( \sqrt{\dfrac{{{u}^{2}}}{11}-1} \right) \right|+C$

Again, we have $u=z-2$ hence the above equation is converted as

$ \int{\dfrac{3{{\left( u+2 \right)}^{3}}-8u-11}{\sqrt{{{u}^{2}}-11}}}={{11}^{\dfrac{3}{2}}}{{\left( \sqrt{\dfrac{{{\left( z-2 \right)}^{2}}}{11}-1} \right)}^{3}}+61\sqrt{11}\left( \sqrt{\dfrac{{{\left( z-2 \right)}^{2}}}{11}-1} \right)+99\left( \dfrac{z-2}{\sqrt{11}} \right)\left( \sqrt{\dfrac{{{\left( z-2 \right)}^{2}}}{11}-1} \right)+112\log \left| \left( \dfrac{z-2}{\sqrt{11}} \right)+\left( \sqrt{\dfrac{{{\left( z-2 \right)}^{2}}}{11}-1} \right) \right|+C $

$ ={{11}^{\dfrac{3}{2}}}{{\left( \dfrac{{{\left( z-2 \right)}^{2}}-11}{11} \right)}^{\dfrac{3}{2}}}+61\sqrt{11}\sqrt{\left( \dfrac{{{\left( z-2 \right)}^{2}}-11}{11} \right)}+99\left( \dfrac{z-2}{\sqrt{11}} \right)\sqrt{\left( \dfrac{{{\left( z-2 \right)}^{2}}-11}{11} \right)}+112\log \left| \left( \dfrac{z-2}{\sqrt{11}} \right)+\sqrt{\left( \dfrac{{{\left( z-2 \right)}^{2}}-11}{11} \right)} \right|+C $

$ ={{\left( {{z}^{2}}-4z-7 \right)}^{\dfrac{3}{2}}}+61\left( {{z}^{2}}-4z-7 \right)+9\left( z-2 \right)\left( {{z}^{2}}-4z-7 \right)+112\log \left| \left( \dfrac{z-2}{\sqrt{11}} \right)+\dfrac{\sqrt{{{z}^{2}}-4z-7}}{\sqrt{11}} \right|+C $

$ ={{\left( \sqrt{{{z}^{2}}-4z-7} \right)}^{3}}+61\left( {{z}^{2}}-4z-7 \right)+9\left( z-2 \right)\left( {{z}^{2}}-4z-7 \right)+112\log \left| z-2+\sqrt{{{z}^{2}}-4z-7} \right|+C $

$ =\left( {{z}^{2}}-4z-7 \right)\left[ {{\left( \sqrt{{{z}^{2}}-4z-7} \right)}^{2}}+61+9z-18 \right]+112\log \left| z-2+\sqrt{{{z}^{2}}-4z-7} \right|+C $

$  =\left( {{z}^{2}}-4z-7 \right)\left[ {{z}^{2}}-4z-7+9z+43 \right]+112\log \left| z-2+\sqrt{{{z}^{2}}-4z-7} \right|+C $

$=\left( {{z}^{2}}-4z-7 \right)\left( {{z}^{2}}+5z+36 \right)+112\log \left| z-2+\sqrt{{{z}^{2}}-4z-7} \right|+C $

Now equating the $L.H.S=R.H.S$ we have

$\left( {{z}^{2}}+5z+36 \right)\left( {{z}^{2}}-4z-7 \right)+112\log \left| z-2+\sqrt{{{z}^{2}}-4z-7} \right|+C=\left( {{z}^{2}}+az+36 \right)\sqrt{{{z}^{2}}-4z-7}+b\log \left| z-2+\sqrt{{{z}^{2}}-4z-7} \right|+C$

From above equation we have

$a=5$ , $b=112$ hence $a+b=117$ and $a < b$ and Exactly one out of $a$ or $b$ is a prime number. Here a is prime number

So, the correct answer is “Option B,C and D”.

Note: In this problem we all need to expand the expressions using algebraic formulas carefully and while finding the integration of the equations having multiple terms it is advisable to split the terms and integrate them individually and then substitute them in the integration equation.