Integration using Partial Fractions MCQ Quiz - Objective Question with Answer for Integration using Partial Fractions - Download Free PDF

Calculation:

We know that the integral is of the form:

$f(x)=\frac{1}{2}{\log }_e\left(\frac{x^2+3}{x^2+1}\right)+C$

Now, we substitute the given value of f(3) to find the constant C :

⇒ $f(3)=\frac{1}{2}{\log }_e\left(\frac{6}{5}\right)+C$

We are given that:

⇒ $f(3)=\frac{1}{2}({\log }_{e}5-{\log }_{e}6)$

Equating the two expressions for f(3):

⇒ $\frac{1}{2}{\log }_e\left(\frac{6}{5}\right)+C=\frac{1}{2}({\log }_{e}5-{\log }_{e}6)$

Since both sides are equal, we conclude that C = 0 .

Thus, the function becomes:

⇒ $f(x)=\frac{1}{2}{\log }_e\left(\frac{x^2+3}{x^2+1}\right)$

Now, we can calculate f(4):

⇒ $f(4)=\frac{1}{2}{\log }_e\left(\frac{16+3}{16+1}\right)=\frac{1}{2}{\log }_e\left(\frac{19}{17}\right)$

Thus, the value of f(4) is:

⇒ $\frac{1}{2}{\log }_e\left(\frac{19}{17}\right)$

 $\frac{1}{2}({\log }_{e}19-{\log }_{e}17)$

Hence, the correct answer is Option 1.

Given:

$\int \frac{2\mathrm{x}+3}{{\mathrm{x}}^3+{\mathrm{x}}^2-2\mathrm{x}}\mathrm{d}\mathrm{x}$

Concept:

Use concept of partial fractions

$\frac{f(x)}{g(x)\cdot h(x)}=\frac{A}{g(x)}+\frac{B}{h(x)}$

And use formula of integration

$\int \frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}=\mathrm{l}\mathrm{o}\mathrm{g}|\mathrm{x}|+\mathrm{c}$

Calculation:

$\int \frac{2\mathrm{x}+3}{{\mathrm{x}}^3+{\mathrm{x}}^2-2\mathrm{x}}\mathrm{d}\mathrm{x}$

$=\int \frac{2\mathrm{x}+3}{\mathrm{x}(\mathrm{x}-1)(\mathrm{x}+2)}\mathrm{d}\mathrm{x}$

Use concept of partial fractions

$\frac{2\mathrm{x}+3}{\mathrm{x}(\mathrm{x}-1)(\mathrm{x}+2)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{\mathrm{x}-1}+\frac{\mathrm{C}}{\mathrm{x}+2}$

Now, Cross multiply by denominators 

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

Compare the coefficients on both the sides.

$\mathrm{A}+\mathrm{B}+\mathrm{C}=0$ .........(1)

$\mathrm{A}+2\mathrm{B}-\mathrm{C}=2$ ...........(2) and

$-2\mathrm{A}=3$

$⟹\mathrm{A}=\frac{-3}{2}$

On adding equation (1) and (2)

$2\mathrm{A}+3\mathrm{B}=2$

Put value of A then

$2\times \frac{-3}{2}+3\mathrm{B}=2$

$⟹\mathrm{B}=\frac{5}{3}$

Now put the value of A and B in equation (1) then we get

$\mathrm{C}=\frac{-1}{6}$

Now put all these values in integral then 

$\int \frac{2\mathrm{x}+3}{{\mathrm{x}}^3+{\mathrm{x}}^2-2\mathrm{x}}\mathrm{d}\mathrm{x}$

$=\int \frac{-3}{2}.\frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}+\int \frac{5}{3}.\frac{1}{\mathrm{x}-1}\mathrm{d}\mathrm{x}+\int \frac{-1}{6}.\frac{1}{\mathrm{x}+2}\mathrm{d}\mathrm{x}$

$=-\frac{3}{2}\mathrm{l}\mathrm{o}\mathrm{g}|\mathrm{x}|+\frac{5}{3}\mathrm{l}\mathrm{o}\mathrm{g}|\mathrm{x}-1|-\frac{1}{6}\mathrm{l}\mathrm{o}\mathrm{g}|\mathrm{x}+2|+\mathrm{c}$

$=\frac{5}{3}\mathrm{l}\mathrm{o}\mathrm{g}|\mathrm{x}-1|-\frac{3}{2}\mathrm{l}\mathrm{o}\mathrm{g}|\mathrm{x}|-\frac{1}{6}\mathrm{l}\mathrm{o}\mathrm{g}|\mathrm{x}+2|+\mathrm{c}$

Hence the option (2) is correct.

Calculation:

Given:

$\int \frac{x}{(x-1)(x-2)}dx$

$\frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$

⇒ $x=A(x-2)+B(x-1)$

$1=A(1-2)+B(1-1)\Rightarrow A=-1$

$2=A(2-2)+B(2-1)\Rightarrow B=2$

The integral becomes:

⇒ $\int (\frac{-1}{x-1}+\frac{2}{x-2})dx$

⇒ $-\int \frac{1}{x-1}dx+2\int \frac{1}{x-2}dx$

⇒ $-\ln |x-1|+2\ln |x-2|+C$

⇒ $\ln \left|\frac{(x-2{)}^2}{x-1}\right|+C$

Hence option 2 is correct

Calculation

Given:

Let $f(x)=\int \frac{x}{(x^2+1)(x^2+3)}dx$

Using partial fraction decomposition:

$\frac{x}{(x^2+1)(x^2+3)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+3}$

Since the numerator is just 'x', B and D must be zero.

$\frac{x}{(x^2+1)(x^2+3)}=\frac{Ax}{x^2+1}+\frac{Cx}{x^2+3}$

Multiplying both sides by $(x^2+1)(x^2+3)$:

$x=Ax(x^2+3)+Cx(x^2+1)$

$x=A{x}^3+3Ax+C{x}^3+Cx$

$x=(A+C)x^3+(3A+C)x$

Comparing coefficients:

$A+C=0⟹C=-A$

$3A+C=1$

Substituting $C=-A$:

$3A-A=1$

$2A=1⟹A=\frac{1}{2}$

$C=-\frac{1}{2}$

$f(x)=\int (\frac{1/2x}{x^2+1}-\frac{1/2x}{x^2+3})dx$

$f(x)=\frac{1}{2}\int \frac{x}{x^2+1}dx-\frac{1}{2}\int \frac{x}{x^2+3}dx$

Let $u=x^2+1⟹du=2xdx$

Let $v=x^2+3⟹dv=2xdx$

$f(x)=\frac{1}{4}\int \frac{du}{u}-\frac{1}{4}\int \frac{dv}{v}$

$f(x)=\frac{1}{4}\ln |u|-\frac{1}{4}\ln |v|+K$

$f(x)=\frac{1}{4}\ln |x^2+1|-\frac{1}{4}\ln |x^2+3|+K$

$f(x)=\frac{1}{4}\ln \left|\frac{x^2+1}{x^2+3}\right|+K$

Given $f(3)=\frac{1}{4}\log \left(\frac{5}{6}\right)$

$f(3)=\frac{1}{4}\ln \left|\frac{3^2+1}{3^2+3}\right|+K$

$f(3)=\frac{1}{4}\ln \left|\frac{10}{12}\right|+K$

$f(3)=\frac{1}{4}\ln \left|\frac{5}{6}\right|+K$

$\frac{1}{4}\log \left(\frac{5}{6}\right)=\frac{1}{4}\log \left(\frac{5}{6}\right)+K⟹K=0$

$f(x)=\frac{1}{4}\ln \left|\frac{x^2+1}{x^2+3}\right|$

Find f(0):

$f(0)=\frac{1}{4}\ln \left|\frac{0^2+1}{0^2+3}\right|$

$f(0)=\frac{1}{4}\ln \left|\frac{1}{3}\right|$

∴ $f(0)=\frac{1}{4}\log \left(\frac{1}{3}\right)$

Hence option 1 is correct

Calculation

Given the integral:

$\int \frac{\log (1+x^4)}{x^3}dx=f(x)\log \left(\frac{1}{g(x)}\right)+{\tan }^{-1}(h(x))+c$

Let $u=\log (1+x^4)$ and $dv=\frac{1}{x^3}dx$.

Then $du=\frac{4x^3}{1+x^4}dx$ and $v=-\frac{1}{2x^2}$.

Using integration by parts:

$\int \frac{\log (1+x^4)}{x^3}dx=uv-\int vdu$

$=-\frac{\log (1+x^4)}{2x^2}-\int (-\frac{1}{2x^2})\frac{4x^3}{1+x^4}dx$

$=-\frac{\log (1+x^4)}{2x^2}+\int \frac{2x}{1+x^4}dx$

Let $x^2=t$, so $2xdx=dt$.

The integral becomes:

$\int \frac{2x}{1+x^4}dx=\int \frac{dt}{1+t^2}={\tan }^{-1}(t)+C={\tan }^{-1}(x^2)+C$

$\int \frac{\log (1+x^4)}{x^3}dx=-\frac{\log (1+x^4)}{2x^2}+{\tan }^{-1}(x^2)+C$

Comparing with $f(x)\log \left(\frac{1}{g(x)}\right)+{\tan }^{-1}(h(x))+c$:

$f(x)=\frac{1}{2x^2}$

$g(x)=1+x^4$

$h(x)=x^2$

$f\left(\frac{1}{x}\right)=\frac{1}{2{\left(\frac{1}{x}\right)}^2}=\frac{x^2}{2}$

$f(x)+f\left(\frac{1}{x}\right)=\frac{1}{2x^2}+\frac{x^2}{2}=\frac{1+x^4}{2x^2}$

$h(x)[f(x)+f\left(\frac{1}{x}\right)]=x^2\left(\frac{1+x^4}{2x^2}\right)=\frac{1+x^4}{2}$

∴ $h(x)[f(x)+f\left(\frac{1}{x}\right)]=\frac{1+x^4}{2}=\frac{g(x)}{2}$

Concept:

Partial Fraction:

Calculation:

Here we have to find the value of $\int \frac{dx}{x\left(x+2\right)}$

Let $\frac{1}{x\left(x+2\right)}=\frac{A}{x}+\frac{B}{x+2}$

⇒ 1 = A (x + 2) + B x --------(1)

By putting x = 0 on both the sides of (1) we get A = 1/2

By putting x = - 2 on both the sides of (1) we get B = - 1/2

$\Rightarrow \frac{1}{x\left(x+2\right)}=\frac{1}{2x}-\frac{1}{2x+4}$

$\Rightarrow \int \frac{dx}{x\left(x+2\right)}=\frac{1}{2}\int \frac{dx}{x}-\frac{1}{2}\int \frac{dx}{x+2}$

As we know that $\int \frac{dx}{x}=\log \left|x\right|+C$  where C is a constant

$\Rightarrow \int \frac{dx}{x\left(x+2\right)}=\frac{1}{2}\log \left|\frac{x}{x+2}\right|+C$where C is a constant

Formula used:

$\int \frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}=\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}+\mathrm{C}$

log_ax - log_ay = $\mathrm{l}\mathrm{o}{\mathrm{g}}_{\mathrm{a}}\frac{\mathrm{x}}{\mathrm{y}}$

Calculation:

$\int \frac{dx}{x(x^2+1)}$

⇒ $\int \frac{\mathrm{d}\mathrm{x}}{\mathrm{x}}-\frac{\mathrm{x}\mathrm{d}\mathrm{x}}{{\mathrm{x}}^2+1}$

⇒ $\int \frac{\mathrm{d}\mathrm{x}}{\mathrm{x}}-\frac{1}{2}\int \frac{2\mathrm{x}\mathrm{d}\mathrm{x}}{{\mathrm{x}}^2+1}$

⇒ ln x - $\frac{1}{2}\mathrm{l}\mathrm{n}|1+{\mathrm{x}}^2|$ + c

⇒ $\frac{1}{2}\ln x^2-$$\frac{1}{2}\mathrm{l}\mathrm{n}|1+{\mathrm{x}}^2|$ + c

⇒ $\frac{1}{2}\mathrm{l}\mathrm{n}\left(\frac{{\mathrm{x}}^2}{{\mathrm{x}}^2+1}\right)+\mathrm{C}$

∴ $\int \frac{dx}{x(x^2+1)}$ is equal to 

$\frac{1}{2}\mathrm{l}\mathrm{n}\left(\frac{{\mathrm{x}}^2}{{\mathrm{x}}^2+1}\right)+\mathrm{C}$.

Concept:

Using partial fraction method

$\frac{1}{(\mathrm{x}-\mathrm{a}).(\mathrm{x}-\mathrm{b})}$ = $\frac{A}{(x-a)}$ + $\frac{B}{(x-b)}$

$\int \frac{\mathrm{d}\mathrm{x}}{\mathrm{x}}=\log \left|\mathrm{x}\right|+\mathrm{c}$

Calculation:

I = $\int \frac{\mathrm{d}\mathrm{x}}{(\mathrm{x}-2)\left(\mathrm{x}-1\right)}$

⇒ $\frac{1}{(\mathrm{x}-2).(\mathrm{x}-1)}$ = $\frac{A}{(x-2)}$ + $\frac{B}{(x-1)}$

⇒ 1 = A(x - 1) + B(x - 2)

Compair cofficient both sides

Cofficient of x is  A + B = 0

Coffiecient of constant 1 = -A - 2B

Solving the equation we get 

A = 1, B = -1

⇒ $\int \frac{\mathrm{d}\mathrm{x}}{(\mathrm{x}-2)\left(\mathrm{x}-1\right)}$ = $\int \frac{dx}{(x-2)}$ - $\int \frac{dx}{(x-1)}$

⇒ log|x - 2| - log|x - 1| + c       

= $\mathrm{l}\mathrm{o}\mathrm{g}\left|\frac{(\mathrm{x}-2)}{(\mathrm{x}-1)}\right|+\mathrm{c}$                             [∵ log m - log n = log($\frac{\mathrm{m}}{\mathrm{n}}$)]

Concept:

Integral property:

• ∫ xn dx = $\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}$+ C ; n ≠ -1
• $\int \frac{1}{\mathrm{x}}\mathrm{d}\mathrm{x}=\ln \mathrm{x}$ + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Calculation:

I = $\int \frac{5\mathrm{x}}{{\mathrm{x}}^2+3\mathrm{x}-4}\mathrm{d}\mathrm{x}$

I = $\int \frac{5\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}+4)}\mathrm{d}\mathrm{x}$

I = $\int \frac{(\mathrm{x}+4)+(4\mathrm{x}-4)}{(\mathrm{x}-1)(\mathrm{x}+4)}\mathrm{d}\mathrm{x}$

I = $\int \frac{1}{\mathrm{x}-1}\mathrm{d}\mathrm{x}+\int \frac{4}{\mathrm{x}+4}\mathrm{d}\mathrm{x}$

I = ln (x - 1) + 4 ln (x + 4) + c

Concept:

• $\int \frac{1}{\mathrm{x}+\mathrm{a}}\mathrm{d}\mathrm{x}=\mathrm{l}\mathrm{o}\mathrm{g}|\mathrm{x}+\mathrm{a}|+\mathrm{C}$
• $\int \frac{1}{(\mathrm{x}+\mathrm{a}{)}^{\mathrm{n}}}\mathrm{d}\mathrm{x}=\frac{1}{(1-\mathrm{n})(\mathrm{x}+\mathrm{a}{)}^{\mathrm{n}+1}}+\mathrm{C}$

Calculation:​

To solve: $\int \frac{2{\mathrm{x}}^3}{({\mathrm{x}}^2+7{)}^2}\mathrm{d}\mathrm{x}$ 

Let us put x² = t ⇒2xdx = dt in $\int \frac{2{\mathrm{x}}^3}{({\mathrm{x}}^2+7{)}^2}\mathrm{d}\mathrm{x}$  

⇒ $\int \frac{2{\mathrm{x}}^3}{({\mathrm{x}}^2+7{)}^2}\mathrm{d}\mathrm{x}=\int \frac{\mathrm{t}}{(\mathrm{t}+7{)}^2}\mathrm{d}\mathrm{t}$

This integrand is a proper rational fraction. therefore, by using the form of partial fraction, we write

⇒ $\frac{\mathrm{t}}{(\mathrm{t}+7{)}^2}=\frac{\mathrm{A}}{\mathrm{t}+7}+\frac{\mathrm{B}}{(\mathrm{t}+7{)}^2}$

⇒ t = At + 7A + B 

By comparing coefficient of t and constant terms on both sides, we get A = 1 and 7A + B = 0

By solving these equation, we get  A = 1 and B = -7

$\Rightarrow \frac{\mathrm{t}}{(\mathrm{t}+7{)}^2}=\frac{1}{\mathrm{t}+7}-\frac{7}{(\mathrm{t}+7{)}^2}$

⇒ $\int \frac{\mathrm{t}}{(\mathrm{t}+7{)}^2}\mathrm{d}\mathrm{t}=\int \frac{1}{\mathrm{t}+7}\mathrm{d}\mathrm{t}-\int \frac{7}{(\mathrm{t}+7{)}^2}\mathrm{d}\mathrm{t}$

⇒ $\int \frac{\mathrm{t}}{(\mathrm{t}+7{)}^2}\mathrm{d}\mathrm{t}=\mathrm{l}\mathrm{o}\mathrm{g}|\mathrm{t}+7|+\frac{7}{(\mathrm{t}+7)}+\mathrm{C}$ 

Now put t = x² in the above equation we get

⇒ $\int \frac{2{\mathrm{x}}^3}{({\mathrm{x}}^2+7{)}^2}\mathrm{d}\mathrm{x}=\mathrm{l}\mathrm{o}\mathrm{g}|{\mathrm{x}}^2+7|+\frac{7}{({\mathrm{x}}^2+7)}+\mathrm{C}$

Concept Used:
${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$
also, sin (π/2 - x) = cos x 
and cos (π/2 - x) = sin x 
$\sin x=\frac{2\tan x/2}{1+{\tan }^{2}x/2},$ $\cos x=\frac{1-{\tan }^{2}x/2}{1+{\tan }^{2}x/2}$

Also, $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|+c$

Calculation:

Let $I={\int }_0^{\pi /2}\frac{{\cos }^{2}x}{\sin x+\cos x}dx\dots (1)$
 ⇒ $I={\int }_0^{\pi /2}\frac{{\cos }^2(\pi /2-x)}{\sin ({\pi }_2-x)+\cos ({\pi }_2-x)}dx$
 ⇒ $I={\int }_0^{\pi /2}\frac{{\sin }^{2}x}{\sin x+\cos x}dx\cdots \text{(2)}$
Adding (1) and (2)
 ⇒ $2I={\int }_0^{\pi /2}\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x+\cos x}dx$
 ⇒ $2I={\int }_0^{\pi /2}\frac{1}{\sin x+\cos x}dx$

Now, let $I_1=\int \frac{1}{\sin x+\cos x}dx$

$\Rightarrow I_1=\int \frac{dx}{\frac{2\tan x/2}{1+{\tan }^{2}x/2}+\frac{1-{\tan }^{2}x/2}{1+{\tan }^{2}x/2}}$
⇒ $I_1=\int \frac{1+{\tan }^{2}x/2}{2\tan x/2+1-{\tan }^{2}x/2}dx$

⇒ $I_1=\int \frac{{\sec }^{2}x/2dx}{2\tan x/2+1-{\tan }^{2}x/2}$

Put tan x/2 = t
 ⇒ $\frac{1}{2}{\sec }^{2}x/2=\frac{dt}{dx}$
 ⇒ ${\sec }^{2}x/2dx=2dt$

⇒ I_{1$=2\int \frac{dt}{2t+1-t^2}$}
⇒ I1 $=-2\int \frac{dt}{t^2-2t-1}$
⇒ I1$=-2\int \frac{dt}{t^2-2t-1+1-1}$
⇒ I1 $=-2\int \frac{dt}{(t-1{)}^2-(\sqrt{2}{)}^2}$
⇒ I1 $=2\int \frac{dt}{(\sqrt{2}{)}^2-(t-1{)}^2}$
 ⇒I1 $=2\frac{1}{2\sqrt{2}}\log \left|\frac{\sqrt{2}+t-1}{\sqrt{2}-t+1}\right|+c$
⇒ I1 $=\frac{1}{\sqrt{2}}\log \left|\frac{\sqrt{2}+\tan x/2-1}{\sqrt{2}-\tan x/2+1}\right|+c$
Using I₁, in I

$\Rightarrow I=\frac{1}{2\sqrt{2}}[\ln \left|\frac{\sqrt{2}+\tan \pi /4-1}{\sqrt{2}-\tan \pi /4+1}\right|-\ln \left|\frac{\sqrt{2}+\tan 0-1}{\sqrt{2}-\tan 0+1}\right|]$
$\Rightarrow I=\frac{1}{2\sqrt{2}}[\ln \left|\frac{\sqrt{2}}{\sqrt{2}}\right|-\ln \left|\frac{\sqrt{2}-1}{\sqrt{2}+1}\right|]$
$\Rightarrow I=\frac{1}{2\sqrt{2}}[\ln \left|\frac{(\sqrt{2}+1{)}^2}{(\sqrt{2}{)}^2-1^2}\right|]$
$\Rightarrow I=\frac{1}{2\sqrt{2}}\ln \left|\frac{2+1+2\sqrt{2}}{2-1}\right|$$\Rightarrow I=\frac{1}{2\sqrt{2}}\ln |3+2\sqrt{2}|$