Integrating functions is one of the core applications of calculus. Sometimes, this is straightforward, as in:

F(x) = ∫( x³ + 8) dx

In a comparatively complicated example of this type, you can use a version of the basic formula for integrating indefinite integrals:

∫ (xⁿ + A)dx = x^{(n + 1)}/(n + 1) + An + C ,

where A and C are constants.

Thus for this example,

∫ x³ + 8 = x⁴/4 + 8x + C.

Integration of Basic Square Root Functions

On the surface, integrating a square root function is awkward. For example, you may be stymied by:

F(x) = ∫ √[(x³) + 2x - 7]dx

But you can express a square root as an exponent, 1/2:

√ x³ = x^3(1/2) = x^(3/2)

The integral therefore becomes:

∫ (x^3/2 + 2x - 7)dx

to which you can apply the usual formula from above:

= x^(5/2)/(5/2) + 2(x²/2) - 7x

= (2/5)x^(5/2) + x² - 7x

Integration of More Complex Square Root Functions

Sometimes, you may have more than one term under the radical sign, as in this example:

F(x) = ∫ [(x + 1)/ √(x - 3)]dx

You can use u-substitution to proceed. Here, you set u equal to the quantity in the denominator:

u = √(x - 3)

Solve this for x by squaring both sides and subtracting:

u² = x - 3

x = u² + 3

This allows you to get dx in terms of u by taking the derivative of x:

dx = (2u)du

Substituting back into the original integral gives

F(x) = ∫ (u² + 3 + 1)/udu

= ∫[(2u³ + 6u + 2u)/u]du

= ∫ (2u² + 8)du

Now you can integrate this using the basic formula and expressing u in terms of x:

∫ (2u² + 8)du = (2/3)u³ + 8u + C

= (2/3) [√(x - 3)]³ + 8[√(x - 3)] + C

= (2/3)(x - 3)^(3/2) + 8(x - 3)^(1/2) + C