11. ⇒ (MHT CET 2023 10th May Evening Shift )

The value of $\int \frac{d x}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}}$ is

A. ${(\frac{x^{4} + 1}{x^{4}})}^{\frac{1}{4}} + c$ , where $c$ is a constant of integration.

B. ${(x^{4} + 1)}^{\frac{1}{4}} + c$ , where $c$ is a constant of integration.

C. $- {(x^{4} + 1)}^{\frac{1}{4}} + c$ , where $c$ is a constant of integration.

D. $- {(\frac{x^{4} + 1}{x^{4}})}^{\frac{1}{4}} + c$ , where $c$ is a constant of integration.

Correct Option is (D)

Let , $I = \int \frac{1}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}} d x = \int \frac{d x}{x^{5} {(1 + \frac{1}{x^{4}})}^{\frac{3}{4}}}$

Put $1 + \frac{1}{x^{4}} = t \Rightarrow \frac{- 4}{x^{5}} d x = dt$

$\begin{aligned} ∴ & = - \frac{1}{4} \int \frac{d t}{t^{\frac{3}{4}}} \\ = - \frac{1}{4} \times 4 t^{\frac{1}{4}} + c = - t^{\frac{1}{4}} + c \\ = - {(1 + \frac{1}{x^{4}})}^{\frac{1}{4}} + c = - {(\frac{x^{4} + 1}{x^{4}})}^{\frac{1}{4}} + c \end{aligned}$

12. ⇒ (MHT CET 2023 10th May Morning Shift )

If $I = \int \frac{d x}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}}$ , then $I$ is

A. ${(\frac{x^{4} + 1}{x})}^{\frac{1}{4}} + c$ , where $c$ is a constant of integration.

B. $\frac{{(x^{4} - 1)}^{\frac{1}{4}}}{x} + c$ , where $c$ is a constant of integration.

C. $- \frac{{(x^{4} + 1)}^{\frac{1}{4}}}{x} + c$ , where $c$ is a constant of integration.

D. $- {(\frac{x^{4} + 1}{x})}^{\frac{1}{4}} + c$ , where $c$ is a constant of integration.

Correct Option is (C)

Let $I = \int \frac{1}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}} d x = \int \frac{d x}{x^{5} {(1 + \frac{1}{x^{4}})}^{\frac{3}{4}}}$

Put $1 + \frac{1}{x^{4}} = t \Rightarrow \frac{- 4}{x^{5}} d x = d t$

$\begin{aligned} ∴ I & = - \frac{1}{4} \int \frac{dt}{t^{\frac{3}{4}}} = - \frac{1}{4} \times 4 t^{\frac{1}{4}} + c = - t^{\frac{1}{4}} + c \\ = - {(1 + \frac{1}{x^{4}})}^{\frac{1}{4}} + c \\ = \frac{- {(x^{4} + 1)}^{\frac{1}{4}}}{x} + c \end{aligned}$

13. ⇒ (MHT CET 2023 9th May Morning Shift )

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

A. $\log | \frac{x e^{x}}{1 + x e^{x}} | + c$ , where $c$ is a constant of integration.

B. $\log | \frac{x e^{x}}{1 + x e^{x}} | - \frac{1}{1 + x e^{x}} + c$ , where c is a constant of integration.

C. $\log | 1 + x e^{x} | + \frac{1}{1 + x e^{x}} + c$ , where $c$ is constant of integration.

D. $\log | \frac{x e^{x}}{1 + x e^{x}} | + \frac{1}{1 + x e^{x}} + c$ , where $c$ is constant of integration.

Correct Option is (D)

$\begin{aligned} Let I = \int \frac{x + 1}{x {(1 + x e^{x})}^{2}} d x = \int \frac{e^{'} (x + 1)}{e^{x} \cdot x {(1 + x e^{x})}^{2}} d x \\ Let x \cdot e^{x} = t \\ \Rightarrow (x + 1) e^{x} d x = dt \\ ∴ I = \int \frac{d t}{t (1 + t)^{2}} \\ = \int \frac{1 + t - t}{t (1 + t)^{2}} d t \\ = \int \frac{1 d t}{t (1 + t)} - \int \frac{1}{(1 + t)^{2}} d t \\ = \int \frac{1 + t - t}{t (1 + t)} - \int \frac{1}{(1 + t)^{2}} d t \\ = \int \frac{1}{t} d t - \int \frac{1}{(t + 1)} d t - \int \frac{1}{(1 + t)^{2}} d t \\ = \log t - \int \frac{dt}{(t + 1)} - \int \frac{1}{(1 + t)^{2}} d dt + c \\ Let y = 1 + t \\ \Rightarrow d y = dt \\ ∴ I = \log t - \int \frac{1}{y} d y - \int \frac{1}{y^{2}} d y + c \\ = \log t - \log y + \frac{1}{y} + c \\ = \log t - \log (1 + t) + \frac{1}{(1 + t)} + c \\ = \log x e^{x} - \log (1 + x e^{x}) + \frac{1}{(1 + x e^{x})} + c \\ ∴ I = \log | \frac{x e^{x}}{1 + x e^{x}} | + \frac{1}{1 + x e^{x}} + c \end{aligned}$

14. ⇒ (MHT CET 2023 9th May Morning Shift )

$\int \frac{e^{\tan^{- 1} x}}{1 + x^{2}} [{(\sec^{- 1} \sqrt{1 + x^{2}})}^{2} + \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})] d x, x > 0 =$

A. ${(\tan^{- 1} x)}^{2} e^{\tan^{- 1} x} + c$ , where $c$ is a constant of integration.

B. $(\tan^{- 1} x) e^{\tan^{- 1} x} + c$ , where $c$ is a constant of integration.

C. $(\tan^{- 1} x) e^{2 \tan^{- 1} x} + c$ , where $c$ is a constant of integration.

D. ${(\tan^{- 1} x)}^{2} e^{2 \tan^{- 1} x} + c$ , where $c$ is a constant of integration.

Correct Option is (A)

$\begin{array}{ll} \int \frac{e^{\tan^{- 1} x}}{1 + x^{2}} [{(\sec^{- 1} \sqrt{1 + x^{2}})}^{2} + \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})] d x, \\ Put x = \tan t \\ ∴ d x = \sec^{2} t d t \end{array}$

$\begin{aligned} ∴ & I = \int \frac{e^{{sen}^{- 1} (\tan t)}}{1 + \tan^{2} t} [{(\sec^{- 1} \sqrt{1 + \tan^{2} t})}^{2} + \cos^{- 1} (\frac{1 - \tan^{2} t}{1 + \tan^{2} t})] \sec^{2} t d t \\ = \int \frac{e^{t}}{\sec^{2} t} [{(\sec^{- 1} (\sec t))}^{2} + \cos^{- 1} (\cos 2 t)] \sec^{2} t d t \\ = \int e^{t} [t^{2} + 2 t] d t \\ = e^{t} \cdot t^{2} + c \dots [\int e^{x} f (x) \cdot f^{'} (x) = e^{x} f (x) + c] \\ = t^{2} \cdot e^{t} + c \\ = {(\tan^{- 1} x)}^{2} e^{\tan^{- 1} x} + c \end{aligned}$

15. ⇒ (MHT CET 2021 21th September Evening Shift )

$\int [\sin | \log x | + \cos | \log x |] d x =$

A. $\sin | \log x | + c$

B. $x \cos | \log x | + c$

C. $\cos | \log x | + c$

D. $x \sin | \log x | + c$

Correct Option is (D)

Let $I = \int [\sin | \log x | + \cos | \log x |] d x$

Put $\log x = t \Rightarrow \frac{1}{x} d x = d t \Rightarrow d x = e^{t} d t$

$\begin{aligned} ∴ I & = \int (\sin + \cos t) e^{t} d t \\ = \int e^{t} (\sin t + \cos t) d t = e^{t} \sin t + c = x \sin | \log x | + c \end{aligned}$

⇒Indefinite Integration