Antiderivatives and u-substitution

The common mistake

Dropping the + C on an indefinite integral, or substituting u but still plugging the old x bounds into the u expression.

Instead: Every indefinite integral ends with + C. With u-substitution on a definite integral, either convert the bounds into u values right away, or switch back to x before plugging in — never mix the two.

Worked example with Guide me

Evaluate the integral of 2x(x² + 1)³ from x = 0 to x = 1.

1. Sugar · Pick u. What's u and du?
   Student · u = x² + 1, du = 2x dx.
2. Sugar · Change the bounds. What is u when x = 0 and x = 1?
   Student · u = 1 and u = 2.
3. Sugar · Integrate u³ from 1 to 2. What do you get?
   Student · u⁴/4: 16/4 − 1/4 = 15/4.

Answer: 15/4.

Quick answers

What is the most common mistake with antiderivatives and u-substitution?

Dropping the + C on an indefinite integral, or substituting u but still plugging the old x bounds into the u expression.

How do I get antiderivatives and u-substitution right?

Every indefinite integral ends with + C. With u-substitution on a definite integral, either convert the bounds into u values right away, or switch back to x before plugging in — never mix the two.

Can you show a worked example?

Evaluate the integral of 2x(x² + 1)³ from x = 0 to x = 1. 15/4.

Related concepts

• Separable differential equations
• The chain rule and implicit differentiation