U-substitution — technique, change of limits for definite integrals: Worked examples

Reminder of the one tool we use (from the parent):

Here $g (x)$ is the inner function we nickname $u$ , and $g^{'} (x)$ is its derivative (its "slope"). See Chain Rule for why the derivative of the inside must appear.

Two antiderivative rules we lean on repeatedly (from Antiderivatives & Indefinite Integrals):

The scenario matrix

Every u-sub problem falls into one of these cells. The last column names the example that covers it — one example per cell, no overlaps.

Cell	What makes it tricky	Covered by
A. Perfect match	derivative of inside sits right there	Ex 1
B1. Positive constant mismatch	you have $x d x$ but need $c x d x$ ( $c > 0$ )	Ex 2
B2. Negative constant mismatch	$g^{'}$ carries a minus sign — you must drag a " $-$ " through	Ex 3
C. Leftover variable	a stray $x$ won't cancel — must solve $x$ in terms of $u$	Ex 4
*D. Definite, limits cross* (decreasing inside)**	inside function decreases → $u (b) < u (a)$ , reversed limits	Ex 5
E. Degenerate limits ( $u (a) = u (b)$ )	new limits are equal → answer is $0$ with no work	Ex 6
F1. Trig inside	inner function is a trig function whose derivative appears	Ex 7
F2. Logarithm $g^{'} / g$ form	numerator is exactly the derivative of the denominator	Ex 8
G. Real-world word problem	you must build the integral first	Ex 9
H. Exam twist (rewrite before you can see $u$ )	no obvious inner function until you split/manipulate	Ex 10

We now walk them in order.

Ex 1 — Cell A: the perfect match

Ex 2 — Cell B1: fix a positive constant

Ex 3 — Cell B2: a NEGATIVE constant mismatch (carry the minus)

Worked example Statement

Evaluate $\int x e^{- x^{2}} d x$ .

Forecast: Inside the exponential is $- x^{2}$ ; its derivative is $- 2 x$ — a negative constant times $x$ . We only have $+ x d x$ . Watch how a " $-$ " sign must be dragged through. Guess whether the answer starts with $+ \frac{1}{2}$ or $- \frac{1}{2}$ .

Choose $u = - x^{2}$ . Why? It is the inner function of $e^{(\cdot)}$ .
Differentiate: $d u = - 2 x d x \Rightarrow x d x = - \frac{1}{2} d u$ . Why? Solve for the chunk we actually have, $x d x$ . Dividing by $- 2$ keeps the minus sign — this is the whole point of Cell B2.
Replace: $\int e^{u} \cdot (- \frac{1}{2}) d u = - \frac{1}{2} \int e^{u} d u$ . Why? $- \frac{1}{2}$ is a constant, so it leaves the integral, carrying its sign.
Integrate: $- \frac{1}{2} e^{u} + C$ . Why? $\int e^{u} d u = e^{u}$ (the exponential is its own antiderivative).
Back-substitute: $- \frac{1}{2} e^{- x^{2}} + C$ .

Verify: $\frac{d}{d x} (- \frac{1}{2} e^{- x^{2}}) = - \frac{1}{2} e^{- x^{2}} \cdot (- 2 x) = x e^{- x^{2}}$ . ✓ The two minus signs multiply to a plus — exactly recovering the integrand.

Ex 4 — Cell C: a stray variable that will not cancel

Worked example Statement

Evaluate $\int x x + 1 d x$ .

Forecast: Inside the root is $x + 1$ ; its derivative is $1$ . But there's an extra lone $x$ multiplying — and $x$ is not a constant, so we cannot pull it out. What do we do with it? Predict: we'll rewrite $x$ using $u$ .

Choose $u = x + 1$ . Why? It's the inside of the root; its derivative $1$ makes $d u = d x$ clean.
Differentiate: $d u = d x$ .
Handle the stray $x$ : from $u = x + 1$ we get $x = u - 1$ . Why this step? The leftover $x$ can't be a constant lifted out (that's the classic mistake). Instead we express every $x$ in terms of $u$ , including this one.
Replace: $\int (u - 1) u d u = \int (u^{3/2} - u^{1/2}) d u$ . Why? Distribute $u = u^{1/2}$ so we integrate powers of $u$ .
Integrate: $\frac{u ^{5/2}}{5/2} - \frac{u ^{3/2}}{3/2} = \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2}$ . Why? Power rule applied term-by-term with $n = \frac{3}{2}$ and $n = \frac{1}{2}$ .
Back-substitute: $\frac{2}{5} (x + 1)^{5/2} - \frac{2}{3} (x + 1)^{3/2} + C$ .

Verify: Differentiate: $\frac{2}{5} \cdot \frac{5}{2} (x + 1)^{3/2} - \frac{2}{3} \cdot \frac{3}{2} (x + 1)^{1/2} = (x + 1)^{3/2} - (x + 1)^{1/2} = (x + 1)^{1/2} [(x + 1) - 1] = x x + 1$ . ✓

Ex 5 — Cell D: definite integral where the limits CROSS (decreasing inside)

Worked example Statement

Evaluate $\int_{0}^{1} \frac{x}{( 4 - x ^{2} ) ^{2}} d x$ .

Forecast: Inside is $u = 4 - x^{2}$ , which decreases as $x$ grows. So as $x$ runs $0 \to 1$ , $u$ runs $4 \to 3$ — the new upper limit ( $3$ ) is below the new lower limit ( $4$ ). The limits literally cross. Watch how we handle a reversed range. Guess a small positive answer.

The figure below plots the inner function $u = 4 - x^{2}$ (pale yellow) across $x \in [0, 1]$ . The blue dot marks the left endpoint $x = 0$ , which sits high at $u = 4$ ; the pink dot marks the right endpoint $x = 1$ , which has slid down to $u = 3$ . The white arrow along the curve shows the sweep downhill as $x$ increases — that is exactly why the new $u$ -limits come out reversed (running $4 \to 3$ ). Why include it? To make "the limits cross" a thing you can see: the endpoints swap their up/down order under the substitution.

Choose $u = 4 - x^{2}$ , $d u = - 2 x d x \Rightarrow x d x = - \frac{1}{2} d u$ . Why? Inside the squared bracket; its derivative $- 2 x$ is proportional to the $x$ in front, with a minus (Cell B2 flavour reused inside a definite integral).
Translate limits (the variable is now $u$ ; the old numbers no longer describe the range — see Definite Integral & Fundamental Theorem of Calculus):
- $x = 0 \Rightarrow u = 4 - 0 = 4$
- $x = 1 \Rightarrow u = 4 - 1 = 3$ So $u$ goes from $4$ down to $3$ — the picture shows this reversed sweep.
Replace (keep the limits in the order they came): $\int_{4}^{3} \frac{1}{u ^{2}} \cdot (- \frac{1}{2}) d u = - \frac{1}{2} \int_{4}^{3} u^{- 2} d u$ . Why? Do not silently flip the limits; write $4 \to 3$ exactly as the substitution produced them.
Flip using the swap rule: $- \frac{1}{2} \int_{4}^{3} = + \frac{1}{2} \int_{3}^{4}$ . Why? $\int_{a}^{b} = - \int_{b}^{a}$ : swapping limits flips the sign, which cancels the leading minus. This is the clean way to deal with crossed limits.
Integrate: $\frac{1}{2} \cdot \frac{u ^{- 1}}{- 1} = - \frac{1}{2 u}$ , evaluated from $3$ to $4$ . Why? Power rule with $n = - 2$ : $\int u^{- 2} d u = \frac{u ^{- 1}}{- 1}$ .
Evaluate: $- \frac{1}{2 \cdot 4} - (- \frac{1}{2 \cdot 3}) = - \frac{1}{8} + \frac{1}{6} = \frac{- 3 + 4}{24} = \frac{1}{24}$ .

Verify: $\frac{1}{24} \approx 0.0417$ — small and positive as forecast; numeric integration of the original from $0$ to $1$ gives $0.041 \overline{6}$ . ✓

Ex 6 — Cell E: degenerate limits ( $u (a) = u (b)$ ), and the geometry of why

Ex 7 — Cell F1: trig inside

Ex 8 — Cell F2: the $g^{'} / g$ logarithm pattern

Worked example Statement

Evaluate $\int \frac{2 x + 1}{x ^{2} + x + 7} d x$ .

Forecast: Look at the bottom $x^{2} + x + 7$ . Its derivative is $2 x + 1$ — exactly the top! Whenever the numerator is the derivative of the denominator, the answer is a natural log. Predict $ln ∣ denominator ∣$ .

Choose $u = x^{2} + x + 7$ . Why? Its derivative sits in the numerator — the special " $\int \frac{g ^{'}}{g} = ln ∣ g ∣$ " signature.
Differentiate: $d u = (2 x + 1) d x$ . Why? The whole numerator times $d x$ becomes $d u$ .
Replace: $\int \frac{1}{u} d u$ . Why? The denominator $x^{2} + x + 7$ becomes $u$ , and the entire numerator $(2 x + 1) d x$ becomes $d u$ — leaving $\frac{d u}{u}$ , a pure $u$ -integral with no $x$ left.
Integrate: $ln ∣ u ∣ + C$ . Why? Log rule — this is the $n = - 1$ case the power rule can't touch, so $\int \frac{1}{u} d u = ln ∣ u ∣$ .
Back-substitute: $ln ∣ x^{2} + x + 7∣ + C$ . (The absolute value is optional here since $x^{2} + x + 7 > 0$ always, but keep it as a habit.)

Verify: $\frac{d}{d x} ln (x^{2} + x + 7) = \frac{2 x + 1}{x ^{2} + x + 7}$ . ✓

Ex 9 — Cell G: a real-world word problem (build then substitute)

Worked example Statement

Water flows into a tank at a rate $r (t) = \frac{6 t}{( t ^{2} + 1 ) ^{2}}$ litres per minute. How much water enters during the first $2$ minutes?

Forecast: "Total amount from a rate" means integrate the rate over the time interval — that's the Definite Integral & Fundamental Theorem of Calculus. We need $\int_{0}^{2} r (t) d t$ . Inner function $t^{2} + 1$ ; derivative $2 t$ ; we have $6 t$ — a constant fix. Guess an answer a bit under $3$ litres.

Set up: $V = \int_{0}^{2} \frac{6 t}{( t ^{2} + 1 ) ^{2}} d t$ . Why? Accumulated volume = integral of inflow rate over the time window (units: $\frac{L}{min} \times min = L$ ).
Choose $u = t^{2} + 1$ , $d u = 2 t d t \Rightarrow 6 t d t = 3 d u$ . Why? Inside of the squared bracket; fix the constant ( $6 t = 3 \cdot 2 t$ ).
Translate limits:
- $t = 0 \Rightarrow u = 0 + 1 = 1$
- $t = 2 \Rightarrow u = 4 + 1 = 5$
Replace: $\int_{1}^{5} \frac{3}{u ^{2}} d u = 3 \int_{1}^{5} u^{- 2} d u$ . Why? The denominator $(t^{2} + 1)^{2}$ becomes $u^{2}$ , and the whole numerator $6 t d t$ becomes $3 d u$ from step 2. The $3$ is a constant, so it may leave the integral (a variable never could) — leaving a clean power of $u$ .
Integrate: $3 \cdot \frac{u ^{- 1}}{- 1} = - \frac{3}{u}$ , from $1$ to $5$ . Why? Power rule with $n = - 2$ .
Evaluate: $- \frac{3}{5} - (- \frac{3}{1}) = - \frac{3}{5} + 3 = \frac{- 3 + 15}{5} = \frac{12}{5} = 2.4 litres$ . Why? Fundamental Theorem: plug the upper $u = 5$ then subtract the lower $u = 1$ into $- \frac{3}{u}$ . The two minus signs on the lower term make it $+ 3$ , and $3 - \frac{3}{5}$ over a common denominator of $5$ gives $\frac{12}{5}$ .

Verify: $2.4$ L is positive, under $3$ as forecast, and units check ( $L/min \cdot min$ ). Numeric integration of $r (t)$ on $[0, 2]$ gives $2.4$ . ✓

Ex 10 — Cell H: the exam twist (rewrite before $u$ appears)

Worked example Statement

Evaluate $\int \frac{1}{x ln x} d x$ (for $x > 1$ ).

Forecast: There's no obvious "inside function times its derivative"... until you notice that $\frac{1}{x}$ is the derivative of $ln x$ . So the awkward-looking $ln x$ is the real inner function. Predict a log-of-a-log.

Rewrite mentally: $\frac{1}{x ln x} = \frac{1}{ln x} \cdot \frac{1}{x}$ . Why this step? Splitting reveals the pair "function $ln x$ and its derivative $\frac{1}{x}$ " — the chain-rule fingerprint that was hidden.
Choose $u = ln x$ , $d u = \frac{1}{x} d x$ . Why? $ln x$ is the inner function; its derivative $\frac{1}{x} d x$ is present.
Replace: $\int \frac{1}{u} d u$ . Why? The factor $\frac{1}{ln x}$ becomes $\frac{1}{u}$ , and the trailing $\frac{1}{x} d x$ becomes $d u$ — so all $x$ -pieces vanish, leaving $\frac{d u}{u}$ .
Integrate: $ln ∣ u ∣ + C$ . Why? Log rule — the $n = - 1$ case again.
Back-substitute: $ln ∣ ln x ∣ + C$ .

Verify: $\frac{d}{d x} ln (ln x) = \frac{1}{l n x} \cdot \frac{1}{x} = \frac{1}{x l n x}$ . ✓

Filling the matrix — a quick recap

Recall Which example killed which cell? (one-to-one)

A perfect match ::: Ex 1 B1 positive constant mismatch ::: Ex 2 B2 negative constant mismatch (carry the minus) ::: Ex 3 C stray variable → solve $x = u - 1$ ::: Ex 4 D definite, limits cross ( $u (b) < u (a)$ ) → flip with $\int_{a}^{b} = - \int_{b}^{a}$ ::: Ex 5 E degenerate equal limits give 0 ::: Ex 6 F1 trig inside ::: Ex 7 F2 logarithm $g^{'} / g$ ::: Ex 8 G real-world (build the integral) ::: Ex 9 H exam twist (rewrite first) ::: Ex 10

Connections

यही cases Hinglish में
Chain Rule — every "perfect match" and "hidden inside" example is this rule run backwards.
Antiderivatives & Indefinite Integrals — why constants (not variables) leave the integral; source of the power and log rules.
Definite Integral & Fundamental Theorem of Calculus — justifies the limit translation and the swap rule $\int_{a}^{b} = - \int_{b}^{a}$ (Ex 5, 6, 9).
Integration by Parts — reach for it when there is no inner-derivative pair.
Trigonometric Substitution — the specialised sub for $a^{2} \pm x^{2}$ forms.