This integral is solved by expanding the numerator, simplifying, and using both the power rule and the rule for integrating 1/x. Convert the radical √x in the numerator to a fractional exponent and convert the fraction 1/x3/2 to a negative exponent x–3/2. Expand (1 + x1/2 )2 and multiply the result by x–3/2. Then solve the three resulting integrals using the rule for integrating 1/x for one term and the power rule for the other two terms.

∫   (1 + √x)2
x3/2		   dx
=  ∫ (1 + x1/2 )2 x–3/2 dx
=  ∫ (1 + 2x1/2 + x) x–3/2 dx
=  ∫ (x–3/2 + 2x–1 + x–1/2) dx
=  ∫ x–3/2 dx   +   2 ∫ x–1 dx   +   ∫ x–1/2 dx
=   1
–1/2		   x–1/2  +  2 ln |x|  +   1
1/2		  x1/2 +  C
=  –2x–1/2  +  2 ln |x|  +  2x1/2  +  C