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

F(x) = ∫( x^{3} + 8) dx

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

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

where A and C are constants.

Thus for this example,

∫ x^{3} + 8 = x^{4}/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) = ∫ √[(x^{3}) + 2x - 7]dx

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

√ x^{3} = 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(x^{2}/2) - 7x

= (2/5)x^{(5/2)} + x^{2} - 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:

u^{2} = x - 3

x = u^{2} + 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) = ∫ (u^{2} + 3 + 1)/udu

= ∫[(2u^{3} + 6u + 2u)/u]du

= ∫ (2u^{2} + 8)du

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

∫ (2u^{2} + 8)du = (2/3)u^{3} + 8u + C

= (2/3) [√(x - 3)]^{3} + 8[√(x - 3)] + C

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