Question

Plese solve d fullowing given down

integration of (e^x + x³)
integration of (√x + 1/√x)

$integration \: of \: \frac{ \cos(2x) }{ \sin(x) + \cos(x) }$


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Answer 1

$\huge\star\large\star\large\star\huge\star$

$\large\star{\int {e}^{x} + {x}^{3} dx}$

${ ={e}^{x} +\frac{{x}^{4}}{4} }$

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$(\int \sqrt{x} + \frac{1}{ \sqrt{x} } ) \\ \\ = \int{x}^{ \frac{1}{2} } + \frac{1}{x} \\ \\ = \frac{ {x}^{ \frac{1}{2} + 1} }{ \frac{1}{2} } + ln(x) + c \\ \\ = \frac{ {x}^{ \frac{3}{2} } }{ \frac{3}{2} } + ln(x) + c \\ \\ = \frac{2 {x}^{ \frac{3}{2} } }{3} + ln(x) + c$

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$\int \frac{ \cos(2x) }{ \sin(x) + \cos(x) } dx \\ \\ = \frac{ {cos}^{2}x - {sin}^{2} x}{ \sin(x) + \cos(x) } \\ \\ = \frac{( \cos(x) + \sin(x) )( \cos(x) - \sin(x)) }{( \sin(x) + \cos(x) )} \\ \\ = \cos(x) + \sin(x)$

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$\huge\star\large\star\large\star\huge\star$

Answer 2

1) refer to the 1st attachment

2)To demonstrate:

∫x1+x−−−−√dx

Substitute u=1+x. compute dudx(1+x)=1, so du=dx. Since we can express x=x+1–1, so we have ∫(u−1)u−−√du

We can express u−−√=u12, so: ∫(u−1)u12du

Distribute the parenthesis: =∫u12u−u12du⇒∫u32−u12du

Use the sum rule: ∫u32du−∫u12du

So now we solve the integral term by term by using the power rule for each term.

Solving for ∫u32du=u32+132+1⇒2u525

Solving for ∫u12du=u12+112+1⇒2u323

Plug in solved integrals: 2u525−2u323

Undo substitution of u=x+1: =2(x+1)525−2(x+1)323

And that is all there is to it. Add the constant of integration to finish the solution:

=2(x+1)525−2(x+1)323+C

3)refer the second attachment