Question

Please help me its urgent​

Answer 1

Answer:

3x^5/3/5+2e^x-logx

Step-by-step explanation:

first do integration of x^2/3

then e^x is constant term

and integration of 1/x is logx

Answer 2

Solution

Given Expression,

$\displaystyle \sf \: l = \int( {x}^{2} - 3x + 4)dx$

We know that,

$\displaystyle \: \sf \: l = \int {x}^{n} dx \\ \\ \longrightarrow \boxed{ \sf \: l = \dfrac{x {}^{n + 1}}{n + 1}}$

Thus,

$\displaystyle \: \sf \: l = \int {x}^{2} dx -3 \int \: xdx + 4 \int \: dx \\ \\ \implies \: \sf \: l = \dfrac{ {x}^{2 + 1} }{2 + 1} - 3 \frac{x {}^{1 + 1} }{1 + 1} + 4 x$

While integrating an indefinite integral,an arbitrary constant 'c' is added to the final step

$\implies \: \boxed{ \boxed{ \sf \: l = \dfrac{x {}^{3} }{3} - \frac{ {3x}^{2} }{2} + 4x + c}}$

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Given Expression,

$\displaystyle \sf \: l = \int( {x}^{ \frac{2}{3} } + 2 {e}^{x} - \dfrac{1}{x} )dx$

• Integral of e^x is e^x + c

Thus,

$\displaystyle \: \sf \: l = \int {x}^{ \frac{2}{3} } dx -2 \int \: {e}^{x} dx + \int {x}^{ - 1} \: dx \\ \\ \implies \: \sf \: l = \dfrac{ {x}^{ \frac{2}{3} + 1} }{ \dfrac{2}{3} + 1} + 2 {e}^{x} - ln(x) \\ \\ \implies \: \boxed{ \boxed{ \sf \: l = \dfrac{3 \sqrt[3]{ {x}^{5} } }{5} + 2 {e}^{x} - log(x) + c}}$

Here,

Integrating x^-1 would result in an undefined term. Therefore,

• Integral of 1/x is log(x) + c

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Given Expression,

$\displaystyle \sf l = \int cosec \ x(cosec \ x - cot \ x ) dx$

The above expression can be re written as :

cosec x ( cosec x - cot x )

= cosec²x - cosec x.cot x

Integrals of :

• cosec²x = - cot x + c

• cosec x.cot x = - cosec x + c

Thus,

$\displaystyle \sf l = \int cosec^2x dx - \int cosec \ x.cot \ xdx \\ \\ \implies \boxed{\boxed{\sf l= cosec \ x - cot \ x + c}}$

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