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Question

23. edr23. xed equalsA) e+c(B) 'er' +C(B)3e

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Solution

The given integration is,

x 2 e x 3 dx

Assume, x 3 =t

Differentiate with respect to x.

3 x 2 = dt dx dx= dt 3 x 2

Substitute the values in integration, we get,

x 2 e x 3 dx = x 2 e t dt 3 x 2 = e t 3 dt = e t 3 +C

Now, substitute the value of t in the above equation, we get,

x 2 e x 3 dx = e x 3 3 +C

Therefore, Option (A) is correct.


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