Integral Calculus jee mains questions | JEE Main Maths Integral Calculus Previous Year Question

Q 51 :

The value of the integral $\int_{- 1}^{1} (\frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1}) d x$ is equal to: [2026]

$3 \log_{e} 2$
$2 \log_{e} 2$
$5 \log_{e} 3$
$3 \log_{e} 3$

1

2

3

4

(2)

$f (x) = \int_{- 1}^{1} (\frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1}) d x$

$f (x) = \int_{- 1}^{1} \frac{x^{3}}{x^{2} + 2 | x | + 1} d x + \int_{}^{} \frac{| x | + 1}{x^{2} + 2 | x | + 1} d x$

Since, $\frac{x^{3}}{x^{2} + 2 | x | + 1}$ is an odd function and $\frac{| x | + 1}{x^{2} + 2 | x | + 1}$ is an even function

$∴ f (x) = 0 + 2 \int_{0}^{1} \frac{| x | + 1}{x^{2} + 2 | x | + 1} d x$

$= 2 \int_{0}^{1} \frac{x + 1}{{(x + 1)}^{2}} d x$

$= 2 \int_{0}^{1} \frac{1}{x + 1} d x = 2 (\log_{e} {(x + 1))}_{0}^{1} = 2 (\log_{e} 2)$

Q 52 :

The value of the integral $\int_{0}^{2} \frac{\sqrt{x (x^{2} + x + 1)}}{(\sqrt{x + 1}) (\sqrt{x^{4} + x^{2} + 1})} d x$ is equal to: [2026]

$\frac{1}{3} \log_{e} (3 - 2 \sqrt{2})$
$\frac{2}{3} \log_{e} (4 + \sqrt{2})$
$\frac{2}{3} \log_{e} (3 + 2 \sqrt{2})$
$\frac{1}{3} \log_{e} (1 + 6 \sqrt{2})$

1

2

3

4

(3)

$\int_{0}^{2} \frac{\sqrt{x (x^{2} + x + 1)}}{\sqrt{x + 1} \sqrt{x^{2} - x + 1} \sqrt{x^{2} + x + 1}} d x$

$= \int_{0}^{2} \sqrt{\frac{x}{x^{3} + 1}} d x$ ... (i)

Let $x^{3 / 2} = t \Rightarrow \frac{3}{2} x^{1 / 2} d x = d t$

$∴ \frac{2}{3} \int_{0}^{2^{3 / 2}} \frac{d t}{\sqrt{t^{2} + 1}}$ [On Putting above values in (i)]

$= \frac{2}{3} (\ln {(t + \sqrt{t^{2} + 1}))}_{0}^{2^{3 / 2}}$

$= \frac{2}{3} \ln (2^{3 / 2} + 3) = \frac{2}{3} \ln (3 + 2 \sqrt{2})$

Q 53 :

Let $f (x) = {\begin{matrix} \frac{1}{3}, & x \leq π / 2 \\ \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}}, & x > π / 2 \end{matrix}$ . If $f$ is continuous at $x = π / 2$ , then the value of $\int_{0}^{3 b - 6}$ $| x^{2} + 2 x - 3 |$ $d x$ is: [2026]

5
2
3
4

1

2

3

4

(4)

$f (x) = {\begin{matrix} 1 / 3, & x \leq \frac{π}{2} \\ \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}}, & x > \frac{π}{2} \end{matrix}$

$\lim_{x \to \frac{π^{-}}{2}} f (x) = \lim_{x \to \frac{π^{+}}{2}} f (x)$

$\Rightarrow \frac{1}{3} = \lim_{x \to \frac{π^{+}}{2}} \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}}$

$= \lim_{h \to 0} \frac{b (1 - \cos h)}{4 h^{2}}$

$= \lim_{h \to 0} \frac{b \cdot 2 \cdot \sin^{2} (\frac{h}{2})}{4 h^{2}} = \frac{b}{8}$

$\Rightarrow \frac{1}{3} = \frac{b}{8} \Rightarrow b = \frac{8}{3}$

$I = \int_{0}^{2}$ $| x^{2} + 2 x - 3 |$ $d x$

$= \int_{0}^{1} - (x^{2} + 2 x - 3) d x + \int_{1}^{2} (x^{2} + 2 x - 3) d x$

$= \frac{5}{3} + \frac{7}{3} = 4$

Q 54 :

If $α = \int_{0}^{2 \sqrt{3}} \log_{2} (x^{2} + 4) d x + \int_{2}^{4} \sqrt{2^{x} - 4} d x$ , then $α^{2}$ is equal to _________ . [2026]

(192) :

We have,

$α = \int_{0}^{2 \sqrt{3}} \log_{2} (x^{2} + 4) d x + \int_{2}^{4} \sqrt{2^{x} - 4} d x$

Let $y = f (x) = \log_{2} (x^{2} + 4) \Rightarrow 2^{y} = x^{2} + 4$

$\Rightarrow x = \sqrt{2^{y} - 4}$

$\Rightarrow f^{- 1} (y) = x$ $[∵ y = f (x)]$

$\Rightarrow f^{- 1} (y) = \sqrt{2^{y} - 4}$

So, $f^{- 1} (x) = \sqrt{2^{x} - 4}$

$∴$ $α = \int_{0}^{2 \sqrt{3}} f (x) d x + \int_{f (0)}^{f (2 \sqrt{3})} f^{- 1} (x) d x$

$= 2 \sqrt{3} f (2 \sqrt{3}) - 0 (f (0))$

$= 2 \sqrt{3} \times 4$ $[∵ f (2 \sqrt{3}) = 4]$

$= 8 \sqrt{3}$

So, $α^{2} = 64 \times 3 = 192$

Q 55 :

If $\int_{π / 6}^{π / 4} (\cot (x - \frac{π}{3}) \cot (x + \frac{π}{3}) + 1) d x = α \log_{e} (\sqrt{3} - 1)$ , then $9 α^{2}$ is equal to _________ . [2026]

(12)

$\int_{π / 6}^{π / 4} {\cot (x - \frac{π}{3}) \cot (x + \frac{π}{3}) + 1} d x$

$= α \ln (\sqrt{3} - 1)$

$Let A = x - \frac{π}{3} and B = x + \frac{π}{3}$

$B - A = (x + \frac{π}{3}) - (x - \frac{π}{3}) = \frac{2 π}{3}$

$\cot (x - \frac{π}{3}) \cot (x + \frac{π}{3}) + 1$

$= \cot (\frac{2 π}{3}) [\cot (x - \frac{π}{3}) - \cot (x + \frac{π}{3})]$ ${Using, \cot A \cdot \cot B + 1 = \cot (B - A) (\cot A - \cot B)}$

$Since, \cot (\frac{2 π}{3}) = \frac{- 1}{\sqrt{3}}$

$I = - \frac{1}{\sqrt{3}} \int_{π / 6}^{π / 4} [\cot (x - \frac{π}{3}) - \cot (x + \frac{π}{3})] d x$

$Using, \int \cot (u) d u = \ln$ $| \sin u |$ $+ c .$

$I = - \frac{1}{\sqrt{3}} {[\ln | \sin (x - \frac{π}{3}) | - \ln | \sin (x + \frac{π}{3}) |]}_{π / 6}^{π / 4}$

$I = - \frac{1}{\sqrt{3}} (\ln (2 - \sqrt{3}) - \ln (\frac{1}{2}))$

$= - \frac{1}{\sqrt{3}} \ln (\frac{2 - \sqrt{3}}{1 / 2})$

$= - \frac{1}{\sqrt{3}} \ln (4 - 2 \sqrt{3})$ $4 - 2 \sqrt{3} = {(\sqrt{3} - 1)}^{2}$

$I = - \frac{1}{\sqrt{3}} \ln {(\sqrt{3} - 1)}^{2} = - \frac{2}{\sqrt{3}} \ln (\sqrt{3} - 1)$

$α = - \frac{2}{\sqrt{3}} \Rightarrow α^{2} = \frac{4}{3}$

$∴$ $9 α^{2} = 9 \times \frac{4}{3} = 12$

Topic Question Set

Q 51 :

The value of the integral $\int_{- 1}^{1} (\frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1}) d x$ is equal to: [2026]

Q 52 :

The value of the integral $\int_{0}^{2} \frac{\sqrt{x (x^{2} + x + 1)}}{(\sqrt{x + 1}) (\sqrt{x^{4} + x^{2} + 1})} d x$ is equal to: [2026]

Q 53 :

Let $f (x) = {\begin{matrix} \frac{1}{3}, & x \leq π / 2 \\ \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}}, & x > π / 2 \end{matrix}$ . If $f$ is continuous at $x = π / 2$ , then the value of $\int_{0}^{3 b - 6}$ $| x^{2} + 2 x - 3 |$ $d x$ is: [2026]

Q 54 :

If $α = \int_{0}^{2 \sqrt{3}} \log_{2} (x^{2} + 4) d x + \int_{2}^{4} \sqrt{2^{x} - 4} d x$ , then $α^{2}$ is equal to _________ . [2026]

Q 55 :

If $\int_{π / 6}^{π / 4} (\cot (x - \frac{π}{3}) \cot (x + \frac{π}{3}) + 1) d x = α \log_{e} (\sqrt{3} - 1)$ , then $9 α^{2}$ is equal to _________ . [2026]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 51 : The value of the integral ∫–11(x3+|x|+1x2+2|x|+1)dx is equal to: [2026]

Q 52 : The value of the integral ∫02x(x2+x+1)(x+1)(x4+x2+1)dx is equal to: [2026]

Q 53 : Let f(x)={13,x≤π/2b(1–sinx)(π–2x)2,x>π/2. If f is continuous at x=π/2, then the value of ∫03b–6|x2+2x–3|dx is: [2026]

Q 54 : If α=∫023log2(x2+4)dx+∫242x-4dx, then α2 is equal to _________ . [2026]

Q 55 : If ∫π/6π/4(cot(x-π3)cot(x+π3)+1)dx=αloge(3-1), then 9α2 is equal to _________ . [2026]

Want Us to Email you About Special Offers & Updates?

Q 51 :

The value of the integral $\int_{- 1}^{1} (\frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1}) d x$ is equal to: [2026]

Q 52 :

The value of the integral $\int_{0}^{2} \frac{\sqrt{x (x^{2} + x + 1)}}{(\sqrt{x + 1}) (\sqrt{x^{4} + x^{2} + 1})} d x$ is equal to: [2026]

Q 53 :

Let $f (x) = {\begin{matrix} \frac{1}{3}, & x \leq π / 2 \\ \frac{b (1 - \sin x)}{{(π - 2 x)}^{2}}, & x > π / 2 \end{matrix}$ . If $f$ is continuous at $x = π / 2$ , then the value of $\int_{0}^{3 b - 6}$ $| x^{2} + 2 x - 3 |$ $d x$ is: [2026]

Q 54 :

If $α = \int_{0}^{2 \sqrt{3}} \log_{2} (x^{2} + 4) d x + \int_{2}^{4} \sqrt{2^{x} - 4} d x$ , then $α^{2}$ is equal to _________ . [2026]

Q 55 :

If $\int_{π / 6}^{π / 4} (\cot (x - \frac{π}{3}) \cot (x + \frac{π}{3}) + 1) d x = α \log_{e} (\sqrt{3} - 1)$ , then $9 α^{2}$ is equal to _________ . [2026]