We need to evaluate
∫4x+32x2+2x+5dx
Consider ∫4x+32x2+2x+5dx=∫4x2x2+2x+5dx+∫32x2+2x+5dx...(1)
Consider ∫42x2+2x+5dx=4∫12x2+2x+5dx
=4∫x(√2x+1√2)2+92dx
Let u=√2x+1√2⇒du=√2dx
=4∫√2u−1√2(2u2+9)du
=4⋅1√2∫√2u−12u2+9du
=4√2(∫√2u2u2+9du−∫12u2+9du)
=4√2(12√2ln|2u2+9|+13√2tan−1(√23u))
Substitute back u=√2x+1√2 we get
=4√2⎛⎝12√2ln∣∣
∣∣2(√2x+1√2)2+9∣∣
∣∣+13√2tan−1(√23⋅(√2x+1√2))⎞⎠
∴∫42x2+2x+5dx=ln|4x2+4x+10|−23tan−1(23x+13)...(2)
Now consider ∫32x2+2x+5dx=3∫12x2+2x+5dx
=3∫1(√2x+1√2)2+92dx
Let u=√2x+1√2⇒du=√2dx
=3∫√22u2+9du
=3√2∫12u2+9du
=3√213√2tan−1(√23u)
Substitute back u=√2x+1√2 we get
=tan−1(√23⋅(√2x+1√2))
∴∫32x2+2x+5dx=3∫12x2+2x+5dx=tan−1(23x+13)...(3)
Substitute (2),(3) in (1) we get
∫4x+32x2+2x+5dx=ln|4x2+4x+10|−23tan−1(23x+13)+tan−1(23x+13)
∴∫4x+32x2+2x+5dx=ln|4x2+4x+10|+13tan−1(23x+13)+c where c is constant of integration.