Question

अ विजिट टू हॉस्पिटल ऐसे​

Answer 1

Answer:

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Explanation:

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Answer 2

Explanation:

Let I=∫

(1+x)

2

xe

x

dx

I=∫

(1+x)

2

(x+1−1)e

x

dx

I=∫

(1+x)

e

x

dx−∫

(1+x)

2

e

x

dx

Applying integration by parts in first integral, we get

I=

(1+x)

e

x

−∫−(

(1+x)

2

1

)e

x

dx−∫

(1+x)

2

e

x

dx+C

I=

(1+x)

e

x

+∫(

(1+x)

2

1

)e

x

dx−∫

(1+x)

2

e

x

dx+C

I=

(1+x)

e

x

+C

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