Antiderivatives: Level 2 Challenges Practice Problems Online

$\large \int \sin (x) \, dx - \int \sin (x ) \, dx = \ ?$

-1 2 An unknown constant 1 0 -2

Determine the indefinite integral of the following expression: $(\tan{x}) ( \tan{x}+ \sec{x} ).$

Details and Assumptions:

Use $C$ as the constant of the integration.

\tan{x} - x + \sin{x} + C

\tan{x} - x + \sec{x} + C

\tan{x} - x + \cos{x} + C

\cos{x} - x + \sec{x} + C

If $f(x)=\displaystyle \int f(x)dx$ ,

then $\displaystyle \int \left( f''(x)+(f'(x))^2+(f(x))^3 \right)dx=0$ will give $\displaystyle \sum_{r=1}^3 \dfrac{(f(x))^r}{r}=?$

If $f\left( \frac{3x-4}{3x+4} \right) = x + 2$ , then what is the antiderivative of $f(x)$ ?

Clarification: Take $C$ as an arbitrary constant.

e^{x+2} \ln(\frac{3x - 4}{3x + 4}) + C

e^{x-2} \ln(\frac{3x - 4}{3x + 4}) + C

\frac{-8}{3}\ln(x-1) + \frac{2x}{3} + C

None of these choices

e^{\frac{3x - 4}{3x + 4}} - \frac{x^2}{2} + C

\frac{8}{3}\ln(x-1) + \frac{2x}{3} + C

e^{\frac{3x + 4}{3x - 4}} - \frac{x^2}{2} + C

If an antiderivative of $f\left( x \right)$ is ${ e }^{ x }$ and that of $g\left( x \right)$ is $\cos { x }$ , then $\displaystyle \int { f\left( x \right) \cos { x } \ dx } +\int { g\left( x \right) { e }^{ x } \ dx }$ is equal to

{ e }^{ x }\cos { x } +C

f\left( x \right) -g\left( x \right) +C

f\left( x \right) +g\left( x \right) +C

f\left( x \right) .g\left( x \right) +C

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Antiderivatives: Level 2 Challenges