1/(^2 x (1 + x)^3) dx = Option 1: (-2)/((1 + x)^2) + C, where C is constant of Integration Option 2: (-4)/((1 + x)^4) + C, where C is constant of Integration Option 3: (-1)/(2(1 + x)^2) + C, where C is constant of Integration Option 4: (1 + x)^3 + C, where C is constant of Integration

Answer: (-1)/(2(1 + x)^2) + C, where C is constant of Integration. Solution: Let u = 1 + x, then du = ^2 x dx = dx/(^2 x). The integral becomes du/u^3 = u^(-3) du = (u^(-2))/(-2) + C = (-1)/(2(1 + x)^2) + C.

Calculus

Integrals

Easy

$\int \frac{1}{\cos^2 x (1 + \tan x)^3} \, dx =$

$\frac{-2}{(1 + \tan x)^2} + C$ , where C is constant of Integration

$\frac{-4}{(1 + \tan x)^4} + C$ , where C is constant of Integration

$\frac{-1}{2(1 + \tan x)^2} + C$ , where C is constant of Integration

$\log(1 + \tan x)^3 + C$ , where C is constant of Integration

✅ Correct Option: 3

Let $u = 1 + \tan x$ , then $du = \sec^2 x \, dx = \frac{dx}{\cos^2 x}$ .

The integral becomes $\int \frac{du}{u^3} = \int u^{-3} \, du = \frac{u^{-2}}{-2} + C = \frac{-1}{2(1 + \tan x)^2} + C$ .

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[where C is an arbitrary constant]

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$\int_2^5 |x - 3|dx$ equals

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$\int \left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\} dx$ is equal to:(where C is an arbitrary constant)

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For $x \in \mathbb{R} - \{-1,0,1\}$ , $\int \frac{1}{x - x^5}dx$ is equal to

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$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{tanx}}$ is equal to

∫1cos⁡2x(1+tan⁡x)3 dx=\int \frac{1}{\cos^2 x (1 + \tan x)^3} \, dx =∫cos2x(1+tanx)31​dx=

Solution

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$\int \frac{1}{\cos^2 x (1 + \tan x)^3} \, dx =$