Let u be the inside, swap dx for du: Substitution simplifies an integral by letting u = the inside function.
Then du = (du/dx) dx, which you use to replace dx.
Integrate in u, then put x back.
IB-style question — first substitution
Find ∫2x(x² + 1)³ dx using u = x² + 1.
Step by step
- u = x²+1 ⇒ du = 2x dx, so 2x dx becomes du.
- Integrate in u, then replace u.
Final answer
(x² + 1)⁴/4 + C.
Pick u so its derivative is present: Choose u so that du (its derivative) already appears as a factor — then the swap is clean.
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Rearrange du to replace the dx-part: Often du gives you part of the integrand.
Rearrange (e.g. x dx = du/2) to replace exactly that part, integrate in u, and substitute back.
IB-style question — a root
Find ∫x√(x² + 4) dx using u = x² + 4.
Step by step
- u = x²+4 ⇒ du = 2x dx ⇒ x dx = ½ du.
- Integrate and replace u.
Final answer
⅓(x² + 4)3/2 + C.
Replace ALL the x's: After substituting, the integral should be entirely in u — no stray x's left before you integrate.
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Convert the x-limits to u-limits: For a definite integral, after choosing u, convert the limits to u-values (substitute each x-limit into u).
Then evaluate entirely in u — no need to switch back to x.
IB-style question — definite substitution
Evaluate ∫₀¹ 2x(x² + 1)³ dx using u = x² + 1.
Step by step
- u = x²+1; limits: x = 0 → u = 1, x = 1 → u = 2; 2x dx = du.
- Evaluate in u.
Final answer
∫₀¹ 2x(x² + 1)³ dx = 15/4.
New limits, no switching back: Once the limits are in u, evaluate directly — you don't convert the answer back to x.