An expression like ∫2x(x²+3)⁴dx can feel overwhelming at first. You could force the parentheses open into a polynomial and grind through it, but that's far too tedious. That's exactly when the technique of substitution comes in handy. If a complicated-looking expression contains "a function paired with its own derivative," you can swap that entire function out for a new variable, u, turning it into a much simpler expression.
In the example above, set u=x²+3 and you get du=2x dx. But look at the original expression — it's multiplied by exactly 2x dx! So the whole expression turns into ∫u⁴du. That's just a plain power integral, giving u⁵/5+C directly, and substituting x²+3 back in for u at the end gives the answer, (x²+3)⁵/5+C. The key is recognizing the pattern of "an inner function" multiplied by "the derivative of that inner function."
The other technique is integration by parts. An expression like ∫x·eˣdx, where two different kinds of functions (a polynomial and an exponential function) are multiplied together, doesn't solve nicely with substitution. In cases like this, you use a formula that runs the product rule for derivatives in reverse: ∫u dv = uv − ∫v du. Split it as u=x, dv=eˣdx, and you get du=dx, v=eˣ. Substitute into the formula, and ∫x·eˣdx = x·eˣ − ∫eˣdx = x·eˣ − eˣ + C. How you split things into u and dv is the key — generally, picking the side that gets simpler when differentiated (the polynomial) as u, and the side that's easy to integrate (an exponential or trig function) as dv, tends to work out well.
How can you be sure these two methods really give the correct answer? The most reliable way is to check it yourself directly. Take a Riemann sum — split the definite integral's interval into very thin (say, 10,000) rectangles and add up all their areas for an approximation — and compare that against the value you get by substituting the interval's two endpoints into the antiderivative found by substitution or integration by parts. If the two values match almost exactly down to the fourth decimal place, that means the antiderivative you found really does compute the correct area.
On our activity page, you can work through ∫2x(x²+a)ⁿdx on the substitution tab and ∫x·eᵏˣdx on the integration-by-parts tab, changing the constants with sliders. It shows the calculation step by step, then places the definite integral value from the formula side by side with the value computed directly from a Riemann sum, so you can see with your own eyes whether they really match.