Integration by substitution

Sometimes it is difficult to integrate a function using the simple formulas of integrals. We have to introduce a second variable to make the integration easier.

Example 1

Calculate ∫(x+1)⁵dx

If it was the integral of x⁵ it would be easy to calculate. We can introduce another variable to make the integral easy to calculate. If we make x + 1 equal to u we will just have to calculate the integral of u. So let's write u = x+1

Then our integral becomes ∫u⁵dx. We calculate the integral as if it was the integral of x⁵.

 ∫u⁵dx = u⁵⁺¹/5+1 + C = 1/6 u⁶ + C

Let's substitute u we get  ∫u⁵dx = 1/6 (x+1)⁶ + C

Example 2

Calculate ∫√4x+3dx

Let's write the expression as a power:

 ∫√4x+3dx = ∫(4x+3)¹/² dx

Let's write u = 4x+3 du = 4dx then dx = du/4. Let's substitute 4x+3 and dx

∫√4x+3dx = ∫u¹/².du/4 = 1/4∫u¹/²du = 1/4(u¹/²+¹/¹/²+¹) + C = 1/4(u³/²/3/2) + C = (1/4)(2/3)u³/² + C = 1/6u³/² + C = 1/6√(4x+3)³ + C

If you are interested in learning more about integrals visit Center for Integral Development