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Question

(2x+3)x2+3x+2dx is equal to.

A

32(x2+3x+2)3/2+ C
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B

32(x2+3x+2)2/3 + C
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C

23(x2+3x+2)2/3 + C
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D

23(x2+3x+2)3/2 + C
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Solution

The correct option is D
23(x2+3x+2)3/2 + C
To solve these types of integrals we express the linear term as a sum of derivative of the quadratic term and a constant term.
Suppose the integral to be I=(ax+b)cx2+dx+edx
Here, (ax+b)=Addx(cx2+dx+e)+B

For this case, (2x+3)=Addx(x2+3x+2)+B
(2x+3)=A(2x+3)+B
Now, comparing the coefficients of x and the constant term we get
A=1 and B=0.
Now we can write our integral as:
I=(2x+3)x2+3x+2dx
I=(ddx(x2+3x+2)} ×x2+3x+2dx

Now we substitute t=(x2+3x+2)
then, dt=(2x+3)dx
substituting these back in the integral we get,
I=t1/2dt
which gives I=t3/23/2=23t3/2+C
substituting back t we get I=23(x2+3x+2)3/2+C


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