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

Calculation:

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

On differentiating both sides, we get

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

\( = e^{-2x} \left[ -2ax^2 + 2(a - b) \right] + (x + b - 2c) \)

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

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

Hence, the correct answer is Option 4.

Calculation:

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

On differentiating both sides, we get

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

\( = e^{-2x} \left[ -2ax^2 + 2(a - b) \right] + (x + b - 2c) \)

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

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

Hence, the correct answer is Option 3.

Calculation:

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

On differentiating both sides, we get

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

\( = e^{-2x} \left[ -2ax^2 + 2(a - b) \right] + (x + b - 2c) \)

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

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

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: 

\(\text{Let } 3 - x^2 = t^2 \)

⇒ \(-2x\, dx = 2t\, dt \)

⇒ \(x\, dx = -t\, dt\)

⇒ \(x^2 = 3 - t^2\)

⇒ \(f(x) = \int x^2 \sqrt{3 - x^2} \cdot x\, dx\)

\(= \int (3 - t^2) \cdot t \cdot (-t\, dt) + c\)

\(= \int (t^4 - 3t^2) dt + c\)

\(= \frac{t^5}{5} - t^3 + c \)

 \(f(x) = \frac{(3 - x^2)^{5/2}}{5} - (3 - x^2)^{3/2} + c\)

⇒ \(f(\sqrt{2}) = \frac{(3 - 2)^{5/2}}{5} - (3 - 2)^{3/2} + c\)

\(= \frac{1^{5/2}}{5} - 1^{3/2} + c = \frac{1}{5} - 1 + c = -\frac{4}{5} \)

⇒ \(\frac{1}{5} - 1 + c = -\frac{4}{5}\)

⇒ \(-\frac{4}{5} + c = -\frac{4}{5}\)

C = 0

\(f(x) = \frac{(3 - x^2)^{5/2}}{5} - (3 - x^2)^{3/2}\)

⇒ \(f(1) = \frac{(3 - 1)^{5/2}}{5} - (3 - 1)^{3/2}\)

\(= \frac{2^{5/2}}{5} - 2^{3/2}\)

\(= 2^{3/2} \left(\frac{2^{2/2}}{5} - 1\right) = 2 \sqrt{2} \left(\frac{2}{5} - 1\right)\)

\(= 2 \sqrt{2} \left(-\frac{3}{5}\right) = - \frac{6 \sqrt{2}}{5}\)

Hence, the Correct answer is Option 4.

Concept:

1 + cos 2x = 2cos² x

1 - cos 2x = 2sin² x

\(\rm \int \cos x\;dx = \sin x + c\)

Calculation:

I = \(\rm \int cos^2 x\;dx\)

= \(\rm \int \frac{1+\cos 2x}{2}\;dx\)

= \(\rm \frac{1}{2}\int (1+\cos 2x)\;dx\)

= \(\rm \frac{1}{2} \left[x+\frac{\sin 2x}{2} \right ] + c\)

= \(\rm \frac{x}{2}+\frac{\sin 2x}{4} + c\)

Concept:

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

Calculation:

I = \(\rm \int \frac {1}{\sqrt{16-25x^2}}dx\)

= \(\rm \int \frac {1}{\sqrt{16-(5x)^2}}dx\)

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = \(\rm \frac {dt}{5}\)

Now,

I = \(\rm \frac {1}{5}\int \frac {1}{\sqrt{4^2-t^2}} dt\)

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

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

Concept:

\(\rm \int x^n dx = \frac{x^{n+1}}{n+1} +c\)

Calculation:

I = \(\rm \int \sqrt{2x+3}\;dx\)

Let 2x + 3 = t²

Differenating with respect to x, we get

⇒ 2dx = 2tdt

⇒ dx = tdt

Now,

I = \(\rm \int \sqrt{t^2}\; \times tdt\)

= \(\rm \int t^2 \;dt\)

= \(\rm \frac {t^3}{3} + c\)

∵ 2x + 3 = t2

⇒  (2x + 3)^1/2 = t

⇒ (2x + 3)³/2 = t³

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

Concept:

\(\rm \int \sin x \; dx = -\cos x + c\)

Calculation:

I = \(\rm \int \sin 5x\;dx \)

Let 5x = t

Differentiating with respect to x, we get

⇒ 5dx = dt

⇒ dx = \(\rm \frac{dt}{5} \)

Now,

I = \(\rm \frac 1 5 \int \sin t\;dt \)

= \(\rm \frac 1 5 (-\cos t) + c\)

= \(\rm \frac{-\cos 5x}{5} + c\)

Concept:

From the Standard integral:

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Calculation:

\(\smallint \frac{{{e^x}}}{{{e^{2x}} - 4}}dx\)

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

let t = e^x

dt = e^x dx

\(\smallint \frac{{{dt}}}{{({t)^2} - (2)^2}}\)

From the standard integral:

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

\(\smallint \frac{{{e^x}}}{{{e^{2x}} - 4}}dx=\frac{1}{4}\log \left| {\frac{{{e^x} - 2}}{{{e^x} + 2}}} \right| + C\)

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

\(\smallint \frac{{{dx}}}{{{\sqrt{{a^2} - x^2}}}}=sin^{-1}(\frac{x}{a})+C\)

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

Concept:

• \(\smallint {\rm{se}}{{\rm{c}}^2}{\rm{x\;dx}} = \tan {\rm{x}} + {\rm{\;C}}\)
• \(\smallint \sec {\rm{x}}\tan {\rm{x\;dx}} = \sec {\rm{x}} + {\rm{c}}\)

Calculation:

Let I = \(\smallint \frac{{\sin {\rm{x}}}}{{{{\left( {\cos {\rm{x}}} \right)}^3}}}{\rm{dx}}\) 

\( = {\rm{\;}}\smallint \tan {\rm{x\;}}{\sec ^2}{\rm{x\;dx}}\)

Let tan x = t

⇒ sec2x dx = dt

Therefore, the integral becomes.

\(= {\rm{\;}}\smallint {\rm{t\;dt}}\)

\( = \frac{{{{\rm{t}}^2}}}{2} + {\rm{\;C}}\)

Re-substitute t = tan x.

Thus,

\( = {\rm{\;}}\frac{{{{\tan }^2}{\rm{x}}}}{2} + {\rm{\;C}}\)

Concept: 

\(\rm \int x^{n}\space dx = \frac{x^{n + 1}}{n + 1} + C\)

\(\rm \int f(x) \space dx^2\) = \(\rm \int (1 + x^{2} + x^{4}) \space d(x^2)\)      ....(i)

Calculation:

Let, x² = u

From equation (i)

​\(\rm \int f(x) \space dx^2\) = \(\rm \int (1 + u + u^{2}) \space du\)

⇒ u + \(\rm \frac{u^{2}}{2}\) + \(\rm \frac{u^{3}}{3}\)+ C

Now putting the value of u,

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

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

Concept:

• \(\smallint {{\rm{x}}^{\rm{n}}}{\rm{dx}} = {\rm{\;}}\frac{{{{\rm{x}}^{\rm{n}+1}}}}{{\rm{n+1}}} + {\rm{c}}\)

Calculation:

Let I = \(\smallint {\sin ^3}x\cos x\;dx\)

Let sin x = t

Now differentiating both sides, we get

⇒ cos x dx = dt

Now,

\({\rm{I\;}} = \smallint {\sin ^3}{\rm{x}}\cos {\rm{x\;dx}} = {\rm{\;}}\smallint {{\rm{t}}^3}{\rm{dt}} = {\rm{\;}}\frac{{{{\rm{t}}^4}}}{4} + {\rm{c}}\)

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

∴ Option 4 is correct answer

Concept:

• cos 2x = cos² x - sin² x

Calculation:

\(\int\frac{\cos2x}{\cos^2x.\sin^2x}dx\)

= \(\int\frac{cos^{2}x-sin^{2}x}{\cos^2x.\sin^2x}dx \)

= \(\int \left ( \frac{1}{sin^{2}x}-\frac{1}{cos^{2}x} \right )dx\)

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

= \(\rm \int cosec^{2}xdx-\int sec^{2}x dx\)

= - cot x - tan x + C

Concept:

\(\rm \displaystyle\int \dfrac{1}{x}dx= \log x + c\)

Calculation:

\(\rm \text{Let I}=\displaystyle\int \dfrac{1}{1+e^x}dx \\=\rm\displaystyle\int \dfrac{e^{-x}}{e^{-x}+1}dx \)

Assume e^-x + 1 = t

Differenatiang with respect to x, we get

⇒ -e^-x dx = dt

∴ e-x dx = -dt

\(\rm =\displaystyle\int \dfrac{-dt}{t}\\= -\log t + c\\=-\log (e^{-x}+1)+c\\=\log \left(\frac{1}{e^{-x}+1} \right )+c\\=\log \left(\frac{e^x}{1+e^x} \right )+c\)