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Question

Evaluate 4x+32x2+2x+5dx.

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Solution

We need to evaluate 4x+32x2+2x+5dx

Consider 4x+32x2+2x+5dx=4x2x2+2x+5dx+32x2+2x+5dx...(1)

Consider 42x2+2x+5dx=412x2+2x+5dx

=4x(2x+12)2+92dx

Let u=2x+12du=2dx

=42u12(2u2+9)du

=4122u12u2+9du

=42(2u2u2+9du12u2+9du)

=42(122ln|2u2+9|+132tan1(23u))

Substitute back u=2x+12 we get

=42122ln∣ ∣2(2x+12)2+9∣ ∣+132tan1(23(2x+12))

42x2+2x+5dx=ln|4x2+4x+10|23tan1(23x+13)...(2)

Now consider 32x2+2x+5dx=312x2+2x+5dx

=31(2x+12)2+92dx

Let u=2x+12du=2dx

=322u2+9du

=3212u2+9du

=32132tan1(23u)

Substitute back u=2x+12 we get

=tan1(23(2x+12))

32x2+2x+5dx=312x2+2x+5dx=tan1(23x+13)...(3)

Substitute (2),(3) in (1) we get

4x+32x2+2x+5dx=ln|4x2+4x+10|23tan1(23x+13)+tan1(23x+13)

4x+32x2+2x+5dx=ln|4x2+4x+10|+13tan1(23x+13)+c where c is constant of integration.

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