Find the area under the curve of f(x)=4x/(x^2+1) form x = 0 to x = 2.

Solution of all integration problems starts with investigation of a given function. We can’t represent it with standard functions, i.e. linear, power or trigonometric. Second step would be to try substitution method, were we substitute f(x) with other function, let’s say u(x). Have the same argument – x, but are different. However, it is not some random function. We divide our initial function f(x) into u(x) and u’(x), new function and new function’s derivative. Essentially, this is called u substitution method.

Now, if you look closely, f(x) consist from linear function (4x) and quadratic (x^2 + 1). Let’s try u substitution, assuming that u(x)= x^2 + 1. Therefore, u’(x) = 2x. 2x is not a 4x, but we could factor out 2, which gives us 2*2x, or 2 * u’(x). Thus, f(x) = (2 * u’(x)) / u(x) = 2 (u’(x) / u(x)). u’(x) is the same as du(x) / dx. 2 ∫(1 / u(x)) * (du(x) / dx) dx. We imagine as dx is divided by dx and we are left with du(x). ∴ 2 ∫1 / u(x) du(x) = 2 * ln(u(x)) + C. Now, substituting u(x) = x^2 + 1 will give us: 2 * ln(x^2 + 1) + C. To find an area under a curve we should subtract integral with x = 0 from integral with x = 2. Area = (2 * ln(2^2 + 1) + C) –(2 * ln(0^2 + 1) + C) = 2 * ln(5) – 2 * ln(1) = 2 * ln(5) And we are done.