This is where most marks are won or lost. The right u makes the integral collapse; the wrong u makes it worse.
The choosing-u checklist (try in this order)
1. Look for a "nest". Something inside brackets, under a root, in an exponent, or inside sin/cos/ln. That inner thing is your first candidate for u.
2. Check its derivative appears elsewhere in the integrand (up to a constant multiple). If yes. You have a winner.
3. Prefer the inside of the "hardest" part. For (x²+1)⁵ , pick u = x²+1, not u = x⁵.
4. For fractions, try u = the denominator (especially if the numerator resembles its derivative → gives ln|u|).
5. Constants don’t matter. If du = 2x dx but you only have x dx , that’s fine: x dx = ½ du . You can always fix a constant, but you can never fix a missing variable.
Pattern-spotting: common setups
Integral looks like
Choose u =
Because
∫ 2x(x²+1)⁵ dx
x² + 1
du = 2x dx: exactly the other factor
∫ x·e^(x²) dx
x²
du = 2x dx, and we have x dx = ½du
∫ cos(3x+1) dx
3x + 1
du = 3 dx, so dx = ⅓du
∫ (2x+3)/(x²+3x) dx
x² + 3x
numerator IS the derivative → ln|u|
∫ sin⁴x · cos x dx
sin x
du = cos x dx: the cos is there for you
∫ x√(x²−4) dx
x² − 4
du = 2x dx; the x outside pairs up
∫ (ln x)²/x dx
ln x
du = (1/x)dx: the 1/x is present
⚠
The golden test
After substituting, your integral must contain no x’s at all, only u and du. If a stray x survives, your choice of u was wrong (or you need to also rearrange u to express that x in terms of u).