Indefinite Integrals MCQ Quiz - Objective Question with Answer for Indefinite Integrals - Download Free PDF

Calculation:

$\int x^2{e}^{-2x}dx=e^{-2x}(a{x}^2+bx+c)+D$

On differentiating both sides, we get

$x^2{e}^{-2x}=e^{-2x}(2ax+b)+(a{x}^2+bx+c)(-2e^{-2x})$

$=e^{-2x}[-2a{x}^2+2(a-b)]+(x+b-2c)$

$\Rightarrow a=1,2(a-b)=0,b-2c=0$

$\Rightarrow a=1,b=1\text{ and }c=\frac{1}{2}$

Hence, the correct answer is Option 4.

Calculation:

$\int x^2{e}^{-2x}dx=e^{-2x}(a{x}^2+bx+c)+D$

On differentiating both sides, we get

$x^2{e}^{-2x}=e^{-2x}(2ax+b)+(a{x}^2+bx+c)(-2e^{-2x})$

$=e^{-2x}[-2a{x}^2+2(a-b)]+(x+b-2c)$

$\Rightarrow a=1,2(a-b)=0,b-2c=0$

$\Rightarrow a=1,b=1\text{ and }c=\frac{1}{2}$

Hence, the correct answer is Option 3.

Calculation:

$\int x^2{e}^{-2x}dx=e^{-2x}(a{x}^2+bx+c)+D$

On differentiating both sides, we get

$x^2{e}^{-2x}=e^{-2x}(2ax+b)+(a{x}^2+bx+c)(-2e^{-2x})$

$=e^{-2x}[-2a{x}^2+2(a-b)]+(x+b-2c)$

$\Rightarrow a=1,2(a-b)=0,b-2c=0$

$\Rightarrow a=1,b=1\text{ and }c=\frac{1}{2}$

Hence, the correct answer is Option 1.

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.

Calculation:

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

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

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

= $\int {\mathrm{e}}^{\mathrm{x}+\frac{1}{\mathrm{x}}}\mathrm{d}\mathrm{x}+\mathrm{x}{\mathrm{e}}^{\mathrm{x}+\frac{1}{\mathrm{x}}}\mathrm{d}\mathrm{x}-\int \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}(\mathrm{x}){\mathrm{e}}^{\mathrm{x}+\frac{1}{\mathrm{x}}}\mathrm{d}\mathrm{x}$

= $\int {\mathrm{e}}^{\mathrm{x}+\frac{1}{\mathrm{x}}}\mathrm{d}\mathrm{x}+\mathrm{x}{\mathrm{e}}^{\mathrm{x}+\frac{1}{\mathrm{x}}}\mathrm{d}\mathrm{x}-\int {\mathrm{e}}^{\mathrm{x}+\frac{1}{\mathrm{x}}}\mathrm{d}\mathrm{x}$ [∵ $\int (1-\frac{1}{{\mathrm{x}}^2}){\mathrm{e}}^{\mathrm{x}+\frac{1}{\mathrm{x}}}\mathrm{d}\mathrm{x}={\mathrm{e}}^{\mathrm{x}+\frac{1}{\mathrm{x}}}$]

= $\mathrm{x}{\mathrm{e}}^{\mathrm{x}+\frac{1}{\mathrm{x}}}+\mathrm{c}$

∴ The value of the integral is $\mathrm{x}{\mathrm{e}}^{\mathrm{x}+\frac{1}{\mathrm{x}}}+\mathrm{c}$.

The correct answer is Option 2.

Concept:

1 + cos 2x = 2cos² x

1 - cos 2x = 2sin² x

$\int \cos \mathrm{x}\mathrm{d}\mathrm{x}=\sin \mathrm{x}+\mathrm{c}$

Calculation:

I = $\int \mathrm{c}\mathrm{o}{\mathrm{s}}^2\mathrm{x}\mathrm{d}\mathrm{x}$

= $\int \frac{1+\cos 2\mathrm{x}}{2}\mathrm{d}\mathrm{x}$

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

= $\frac{1}{2}[\mathrm{x}+\frac{\sin 2\mathrm{x}}{2}]+\mathrm{c}$

= $\frac{\mathrm{x}}{2}+\frac{\sin 2\mathrm{x}}{4}+\mathrm{c}$

Concept:

$\int \frac{1}{\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}\mathrm{d}\mathrm{x}={\sin }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c}$

Calculation:

I = $\int \frac{1}{\sqrt{16-25{\mathrm{x}}^2}}\mathrm{d}\mathrm{x}$

= $\int \frac{1}{\sqrt{16-(5\mathrm{x}{)}^2}}\mathrm{d}\mathrm{x}$

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = $\frac{\mathrm{d}\mathrm{t}}{5}$

Now,

I = $\frac{1}{5}\int \frac{1}{\sqrt{4^2-{\mathrm{t}}^2}}\mathrm{d}\mathrm{t}$

= $\frac{1}{5}{\sin }^{-1}\left(\frac{\mathrm{t}}{4}\right)$ + c

= $\frac{1}{5}{\sin }^{-1}\left(\frac{5\mathrm{x}}{4}\right)$ + c

Concept:

$\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$

Calculation:

I = $\int \sqrt{2\mathrm{x}+3}\mathrm{d}\mathrm{x}$

Let 2x + 3 = t²

Differenating with respect to x, we get

⇒ 2dx = 2tdt

⇒ dx = tdt

Now,

I = $\int \sqrt{{\mathrm{t}}^2}\times \mathrm{t}\mathrm{d}\mathrm{t}$

= $\int {\mathrm{t}}^2\mathrm{d}\mathrm{t}$

= $\frac{{\mathrm{t}}^3}{3}+\mathrm{c}$

∵ 2x + 3 = t2

⇒  (2x + 3)^1/2 = t

⇒ (2x + 3)³/2 = t³

⇒ I = $\frac{(2\mathrm{x}+3{)}^{3/2}}{3}+\mathrm{c}$

Concept:

$\int \sin \mathrm{x}\mathrm{d}\mathrm{x}=-\cos \mathrm{x}+\mathrm{c}$

Calculation:

I = $\int \sin 5\mathrm{x}\mathrm{d}\mathrm{x}$

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = $\frac{\mathrm{d}\mathrm{t}}{5}$

Now,

I = $\frac{1}{5}\int \sin \mathrm{t}\mathrm{d}\mathrm{t}$

= $\frac{1}{5}(-\cos \mathrm{t})+\mathrm{c}$

= $\frac{-\cos 5\mathrm{x}}{5}+\mathrm{c}$

Concept:

From the Standard integral:

$$$\int \frac{dx}{x^2-a^2}=\frac{1}{2a}log|\frac{x-a}{x+a}|+C,x>a$

Calculation:

$\int \frac{e^x}{e^{2x}-4}dx$

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

let t = e^x

dt = e^x dx

$\int \frac{dt}{(t{)}^2-(2{)}^2}$

From the standard integral:

$$$\int \frac{dt}{(t{)}^2-(2{)}^2}=\frac{1}{4}log|\frac{t-a}{t+a}|+C$

Put t = ex in the above equation, we get:

$\int \frac{e^x}{e^{2x}-4}dx=\frac{1}{4}\log \left|\frac{e^x-2}{e^x+2}\right|+C$

Note:

Some important formulas of integration are:

\(\smallint \frac{{{dx}}}{{{a^2} - x^2}}=\frac{1}{2a}log \ |\frac{a+x}{a-x}|+C, \ x

$\int \frac{dx}{\sqrt{a^2-x^2}}=si{n}^{-1}(\frac{x}{a})+C$

$\int \frac{dx}{\sqrt{x^2+a^2}}=log(x+\sqrt{a^2+x^2})+C$

Concept:

• $\int \mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}\mathrm{d}\mathrm{x}=\tan \mathrm{x}+\mathrm{C}$
• $\int \sec \mathrm{x}\tan \mathrm{x}\mathrm{d}\mathrm{x}=\sec \mathrm{x}+\mathrm{c}$

Calculation:

Let I = $\int \frac{\sin \mathrm{x}}{{\left(\cos \mathrm{x}\right)}^3}\mathrm{d}\mathrm{x}$ 

$=\int \tan \mathrm{x}{\sec }^2\mathrm{x}\mathrm{d}\mathrm{x}$

Let tan x = t

⇒ sec2x dx = dt

Therefore, the integral becomes.

$=\int \mathrm{t}\mathrm{d}\mathrm{t}$

$=\frac{{\mathrm{t}}^2}{2}+\mathrm{C}$

Re-substitute t = tan x.

Thus,

$=\frac{{\tan }^2\mathrm{x}}{2}+\mathrm{C}$

Concept: 

$\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}$

$\int \mathrm{f}(\mathrm{x})\mathrm{d}{\mathrm{x}}^2$ = $\int (1+{\mathrm{x}}^2+{\mathrm{x}}^4)\mathrm{d}({\mathrm{x}}^2)$      ....(i)

Calculation:

Let, x² = u

From equation (i)

​$\int \mathrm{f}(\mathrm{x})\mathrm{d}{\mathrm{x}}^2$ = $\int (1+\mathrm{u}+{\mathrm{u}}^2)\mathrm{d}\mathrm{u}$

⇒ u + $\frac{{\mathrm{u}}^2}{2}$ + $\frac{{\mathrm{u}}^3}{3}$+ C

Now putting the value of u,

​⇒ $\int \mathrm{f}(\mathrm{x})\mathrm{d}{\mathrm{x}}^2$ = x² +​ $\frac{{\mathrm{x}}^4}{2}$ + $\frac{{\mathrm{x}}^6}{3}$ + C

∴ The required integral is x2 +​ $\frac{{\mathrm{x}}^4}{2}$ + $\frac{{\mathrm{x}}^6}{3}$ + C.

Concept:

• $\int {\mathrm{x}}^{\mathrm{n}}\mathrm{d}\mathrm{x}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$

Calculation:

Let I = $\int {\sin }^{3}x\cos xdx$

Let sin x = t

Now differentiating both sides, we get

⇒ cos x dx = dt

Now,

$\mathrm{I}=\int {\sin }^3\mathrm{x}\cos \mathrm{x}\mathrm{d}\mathrm{x}=\int {\mathrm{t}}^3\mathrm{d}\mathrm{t}=\frac{{\mathrm{t}}^4}{4}+\mathrm{c}$

$\Rightarrow \mathrm{I}=\frac{{\sin }^4\mathrm{x}}{4}+\mathrm{c}=\frac{{\left(1-{\cos }^2\mathrm{x}\right)}^2}{4}+\mathrm{c}$

∴ Option 4 is correct answer

Concept:

• cos 2x = cos² x - sin² x

Calculation:

$\int \frac{\cos 2x}{{\cos }^{2}x.{\sin }^{2}x}dx$

= $\int \frac{co{s}^{2}x-si{n}^{2}x}{{\cos }^{2}x.{\sin }^{2}x}dx$

= $\int (\frac{1}{si{n}^{2}x}-\frac{1}{co{s}^{2}x})dx$

= $\int \frac{1}{{\sin }^{2}x}dx-\int \frac{1}{{\cos }^{2}x}dx$

= $\int \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}\mathrm{d}\mathrm{x}-\int \mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}\mathrm{d}\mathrm{x}$

= - cot x - tan x + C

Concept:

${\displaystyle \int {\displaystyle \frac{1}{\mathrm{x}}}\mathrm{d}\mathrm{x}=\log \mathrm{x}+\mathrm{c}}$

Calculation:

$$\begin{gathered} \text{Let I}={\displaystyle \int {\displaystyle \frac{1}{1+{\mathrm{e}}^{\mathrm{x}}}}\mathrm{d}\mathrm{x}} \\ ={\displaystyle \int {\displaystyle \frac{{\mathrm{e}}^{-\mathrm{x}}}{{\mathrm{e}}^{-\mathrm{x}}+1}}\mathrm{d}\mathrm{x}} \end{gathered}$$

Assume e^-x + 1 = t

Differenatiang with respect to x, we get

⇒ -e^-x dx = dt

∴ e-x dx = -dt

$$\begin{gathered} ={\displaystyle \int {\displaystyle \frac{-\mathrm{d}\mathrm{t}}{\mathrm{t}}}} \\ =-\log \mathrm{t}+\mathrm{c} \\ =-\log ({\mathrm{e}}^{-\mathrm{x}}+1)+\mathrm{c} \\ =\log \left(\frac{1}{{\mathrm{e}}^{-\mathrm{x}}+1}\right)+\mathrm{c} \\ =\log \left(\frac{{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{\mathrm{x}}}\right)+\mathrm{c} \end{gathered}$$