Reverse a chain rule without losing the bounds

Try

Differentiate (x²+1)^4. The result contains both an outer power and the inner derivative 2x. Now examine ∫2x(x²+1)³dx. Its structure suggests reversing that chain rule. Before choosing a substitution, identify the complete inner expression and the factor that represents its derivative.

Predict

If u=x²+1, predict the u values corresponding to x=0 and x=1. Should the transformed integral retain bounds 0 and 1? Those numbers originally measured x, not u. A correct symbolic substitution can still give a wrong answer if its endpoints refer to the wrong variable.

Experiment

Transform ∫₀¹2x(x²+1)³dx step by step.

Original object	Transformed object	Reason
x²+1	u	chosen inner quantity
2x dx	du	derivative of the inner quantity
Lower x=0	u=1	evaluate x²+1
Upper x=1	u=2	evaluate x²+1

The new integral is ∫₁²u³du=[u⁴/4]₁²=15/4. Independently, the original primitive is (x²+1)^4/4; evaluating it at x endpoints gives the same result. These are two consistent bookkeeping routes. Mixing u primitives with x endpoints is not a third route.

Understand

Substitution reverses a chain derivative. If F′=f and g has appropriate regularity, the derivative of F(g(x)) is f(g(x))g′(x). Therefore an integral of that product can be evaluated using F(g(x)), or transformed into the integral of f(u) with u=g(x) and matching endpoint values. The notation du=g′(x)dx records this change; it is not permission to cancel arbitrary symbols without a valid relationship.

For ∫x/(1+x²)dx, choose u=1+x². Since du=2x dx, the original x dx becomes du/2. The result is (1/2)ln(1+x²)+C. Differentiation verifies both the substitution and the factor. Omitting 1/2 doubles the derivative and reveals the error immediately.

For ∫₀³1/√(x+1)dx, choose u=x+1 with bounds 1 and 4. Its primitive is 2√u, giving 2. Domain checks remain important: the original interval stays inside the square root’s allowed region and does not encounter the denominator’s zero.

A substitution does not always require the inner function to be one-to-one when it is justified directly as a chain-rule primitive. For ∫₋₁¹2x cos(x²)dx, the primitive sin(x²) has equal endpoint values, giving zero. As x moves through the interval, u=x² travels down to zero and back up; the signed contributions cancel. This does not justify ignoring branches in arbitrary changes of variables involving inverse expressions or radicals. Those require a range and orientation analysis.

A useful choice makes the remaining integrand entirely expressible in the new variable. If unexplained x terms remain, either solve for them with correct branches or reconsider the substitution. Sometimes expanding an elementary polynomial is simpler and more transparent than introducing a new variable.

Apply

Use the practice problems to identify a missing factor, repair incorrect bounds and explain a nonmonotone inner function. Show the derivative check for each proposed indefinite primitive. Written reasoning should distinguish why a transformation works from merely naming u.

Continue

Next use integrals to measure regions. Correct antiderivatives cannot rescue an incorrectly chosen top boundary, intersection interval or slicing direction.