(√( x))/( x x) dx equals : Option 1: 2√( x) + c Option 2: 2√( x) + c Option 3: √( x) + c Option 4: 2/(√( x)) + c

Answer: 2√( x) + c. Solution: Rewrite: (√( x))/( x x)dx = (^2 x)/(√( x))dx. Let u = x, du = ^2 x dx. Integral becomes du/√u = 2√u + c = 2√( x) + c.

Calculus

Integrals

Medium

core

$\int \frac{\sqrt{\tan x}}{\sin x \cos x} dx$ equals :

$2\sqrt{\tan x} + c$

$2\sqrt{\cot x} + c$

$\sqrt{\tan x} + c$

$\frac{2}{\sqrt{\tan x}} + c$

✅ Correct Option: 1

Rewrite: $\int \frac{\sqrt{\tan x}}{\sin x \cos x}dx = \int \frac{\sec^2 x}{\sqrt{\tan x}}dx$ . Let $u = \tan x$ , $du = \sec^2 x\,dx$ . Integral becomes $\int \frac{du}{\sqrt{u}} = 2\sqrt{u} + c = 2\sqrt{\tan x} + c$ .

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$\int \frac{(x-1)e^x}{x^2} dx, x > 0$ equals (where C is an arbitrary constant)

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$\int e^x \left(\frac{1-x}{1+x^2}\right)^2 dx$ is equal to

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$\displaystyle\int e^x(\sec x + \log_e(\sec x + \tan x))dx$ is equal to: (where C is an arbitrary constant)

∫tan⁡xsin⁡xcos⁡xdx\int \frac{\sqrt{\tan x}}{\sin x \cos x} dx∫sinxcosxtanx​​dx equals :

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$\int \frac{\sqrt{\tan x}}{\sin x \cos x} dx$ equals :