How to Integrate Square Root Functions

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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 = x3(1/2) = x(3/2)

The integral therefore becomes:

∫ (x3/2 + 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

References

About the Author

Kevin Beck holds a bachelor's degree in physics with minors in math and chemistry from the University of Vermont. Formerly with ScienceBlogs.com and the editor of "Run Strong," he has written for Runner's World, Men's Fitness, Competitor, and a variety of other publications. More about Kevin and links to his professional work can be found at www.kemibe.com.

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