Integrating Square Root Functions

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

(F(x) = int( 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:

(int (x^n + A) dx = frac{x^{(n + 1)}}{n + 1} + Ax + C)

where A and C are constants.

Thus for this example,

(int x^3 + 8 = frac{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) = int sqrt{(x^3) + 2x – 7}dx)

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

(sqrt{x^3} = x^{3(1/2)} = x^{(3/2)})

The integral therefore becomes:

(int (x^{3/2} + 2x – 7)dx)

to which you can apply the usual formula from above:

(begin{aligned}int (x^{3/2} + 2x – 7)dx &= frac{x^{(5/2)}}{5/2} + 2bigg(frac{x^2}{2}bigg) – 7x)(&= frac{2}{5}x^{(5/2)} + x^2 – 7xend{aligned})

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

(F(x) = int frac{x + 1}{sqrt{x – 3}}dx)

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

(u = sqrt{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

(begin{aligned})(F(x) &= int frac{u^2 + 3 + 1}{u}(2u)du)(&= int frac{2u^3 + 6u + 2u}{u}du)(&= int (2u^2 + 8)du)(end{aligned})

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

(begin{aligned})(int (2u^2 + 8)du &= frac{2}{3}u^3 + 8u + C)(&= frac{2}{3} (sqrt{x – 3})^3 + 8( sqrt{x – 3}) + C)(&= frac{2}{3} (x – 3)^{(3/2)} + 8(x – 3)^{(1/2)} + C)(end{aligned})