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Question

Find : (x+3)34xx2dx.


Solution

Let I = (x+3)34xx2dx.

Consider x + 3 = A ddx[34xx2]+B     x+3=A[42x]+B

Comparing the like terms, we get A = 12, B = 1,

Therefore, I=12(42x)34xx2 dx+34xx2 dx

Put 34xx2=t (-4-2x)dx =dt in the 1st integral.

So, I=12t dt+8(x2+4x3) dxI=12×23(t)32+((x+2)27) dx

I=13(t)32+(72)(x+2)2dx

I=13(t)32+[(x+2)234xx2+72sin1(x+2)7]+CI=13(34xx2)32+[(x+2)234xx2+72sin1(x+2)7]+C.


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