Question

integrate e^x-sinx+x4​

Answer 1

Answer:

Now,

∫e

x

(sinx+cosx)dx

=∫e

x

sinx dx+∫e

x

cosx dx

=e

x

(sinx)−∫(cosx).e

x

dx+∫e

x

cosx dx

=e

x

(sinx)+c [ Where c is integrating constant]

Step-by-step explanation:

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Answer 2

Answer:$e^{x}+cosx+2x^{2}$

Explanation: Let $I=$$e^x-sinx+4x$.........(1)

From the basic rules of Integration in mathematics  

We know that the,

Integration of $e^{x}= e^{x}$......(2)

Also, we know that the Integration of $sinx=-cosx$....(3)

and Integration of $x=x^{2}/2$.....(4)

Integrating both sides with respect to x in (1)

We got,

I=∫$(e^{x}-sinx+4x)$...(5)

Put the known values of basic rules of Integration in mathematics from (2),(3), (4) in(5)

Hence we get,

I= $e^{x}-(-cosx)+4(x^{2})/2$

On solving we get,

I=$e^{x}+cosx+2x^{2}$.

Hence the required answer of the integration is $e^{x}+cosx+2x^{2}$

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