01 · Rewrite the outside factor
With u = x + 5, rewrite ∫x√(x + 5) dx in u.
Hint
x = u − 5.
Worked solution
∫(u − 5)u¹ᐟ² du = ∫(u³ᐟ² − 5u¹ᐟ²) du.
Understand · explore · practise
Integration by substitution
Learn integration by substitution with a manual change-of-variable model. Rewrite every x and dx, handle roots and exponentials, reverse the substitution and check original worked examples.
Before you startThe chain rule, standard integrals, algebraic simplification and logarithmic integrals.
01 / Change the variable consistently
A substitution reorganises the integral into a simpler one.
If u = g(x), then du = g′(x) dx
This differential notation packages the chain rule; rewrite the complete integral consistently.
Choose a new variable u, find its derivative, and rewrite both the integrand and dx in terms of u. Integrate the simpler expression, then substitute back for an indefinite integral. Leaving some x terms behind usually means the change is incomplete.
Change every part of the integralExplore
Start with the integral in x. Choose u to simplify its repeated expression.
At x = 1: u = 3; du/dx = 1.
Original integrand = 1.732051; transformed integrand × du/dx = 1.732051.
After integrating, replace u with x + 2.
The numerical check illustrates the identity at one input; the symbolic substitution establishes it for the whole valid interval.
02 / Rewrite a leftover x
A linear shift changes the outside factor as well.
Find ∫x√(x + 2) dx using u = x + 2.Worked example
x = u − 2; dx = du
Both the leftover x and square root must change.
∫(u − 2)u¹ᐟ² du
The integral is now entirely in u.
∫(u³ᐟ² − 2u¹ᐟ²) du
Expand before integrating.
(2/5)u⁵ᐟ² − (4/3)u³ᐟ² + C
Use the power rule.
(2/5)(x + 2)⁵ᐟ² − (4/3)(x + 2)³ᐟ² + C
Return to x; use x > −2 for the smooth real interval.
Watch: track each factor through a substitution
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Carry the differential factor
dx does not always become du.
Find ∫x(2x + 3)² dx using u = 2x + 3.Worked example
x = (u − 3)/2; dx = du/2
There are two separate factors of 1/2.
(1/4)∫(u − 3)u² du
One factor comes from x, the other from dx.
u⁴/16 − u³/4 + C
Expand u³ − 3u² and integrate.
(2x + 3)⁴/16 − (2x + 3)³/4 + C
Substitute back.
02 · Find both factors
With u = 3x + 1, rewrite x dx.
Hint
Write x and dx separately.
Worked solution
x = (u − 1)/3 and dx = du/3, so x dx = (u − 1)du/9.
03 · A direct chain substitution
Find ∫6x(x² + 4)² dx.
Hint
Set u = x² + 4, so 6x dx = 3du.
Worked solution
∫3u² du = u³ + C = (x² + 4)³ + C.
04 / Simplify before integrating
The transformed integral may need expansion or division.
Find ∫x/(x + 3) dx using u = x + 3.Worked example
x = u − 3; dx = du
Replace every x.
∫(u − 3)/u du = ∫(1 − 3/u) du
Divide each term by u.
u − 3ln|u| + C
Remember the logarithmic term.
x + 3 − 3ln|x + 3| + C
Equivalent to x − 3ln|x + 3| + C after absorbing 3 into C.
04 · A shifted reciprocal
Find ∫x/(x + 2)² dx.
Hint
With u = x + 2, the integrand becomes u⁻¹ − 2u⁻².
Worked solution
ln|x + 2| + 2/(x + 2) + C, on intervals avoiding x = −2.
05 · Absorbing a constant
Do x + 3 − 3ln|x + 3| + C and x − 3ln|x + 3| + D describe the same primitive family?
Hint
C and D are arbitrary constants.
Worked solution
Yes: take D = C + 3 on the same domain interval.
05 / Remove a square root using u = √x
Express x and dx through the new variable.
Find ∫1/(1 + √x) dx for x > 0.Worked example
u = √x, so x = u² and dx = 2u du
The new variable satisfies u > 0.
∫2u/(1 + u) du
Do not lose the differential’s 2u factor.
∫[2 − 2/(1 + u)] du
Use 2u = 2(1 + u) − 2.
2u − 2ln(1 + u) + C
The logarithm argument is positive.
2√x − 2ln(1 + √x) + C
Return to x and differentiate to check on x > 0.
06 · Root substitution ledger
For u = √(x + 1), write x and dx in terms of u.
Hint
Square both sides first.
Worked solution
x = u² − 1 and dx = 2u du, taking u > 0 on x > −1.
07 · A cancellation after substitution
Find ∫1/[√x(1 + √x)] dx for x > 0.
Hint
Set u = √x and dx = 2u du.
Worked solution
∫2/(1 + u) du = 2ln(1 + √x) + C.
06 / Turn exponentials into powers of u
If u = eˣ, then dx = du/u.
Find ∫e²ˣ/(1 + eˣ) dx.Worked example
u = eˣ > 0; e²ˣ = u²; dx = du/u
Transform the numerator and differential.
∫u/(1 + u) du
Cancel one u, valid because u > 0.
∫[1 − 1/(1 + u)] du
Divide before integrating.
eˣ − ln(1 + eˣ) + C
Integrate in u, then return to x.
08 · Check an exponential differential
If u = e²ˣ, what is dx?
Hint
du/dx = 2e²ˣ = 2u.
Worked solution
dx = du/(2u).
09 · Exponential quotient
Find ∫eˣ/(2 + eˣ) dx.
Hint
Use u = eˣ or directly use g = 2 + eˣ.
Worked solution
ln(2 + eˣ) + C.
07 / Track a negative differential
A cosine substitution often introduces a minus sign.
Find ∫sin x/(2 + cos x)² dx.Worked example
u = 2 + cos x; du = −sin x dx
Keep the negative sign.
−∫u⁻² du
The whole integrand is in u.
u⁻¹ + C = 1/(2 + cos x) + C
The integration sign cancels the negative exponent’s sign.
Derivative = sin x/(2 + cos x)²
Check the final answer directly; the denominator is positive.
10 · Sine as the new variable
Find ∫cos x√(3 + sin x) dx.
Hint
Use u = 3 + sin x.
Worked solution
(2/3)(3 + sin x)³ᐟ² + C.
08 / Rewrite a nonlinear leftover factor
Separate a useful differential from what remains.
Find ∫x³/(x² + 1) dx using u = x² + 1.Worked example
x³ dx = x²(x dx)
Keep x dx together because du = 2x dx.
x² = u − 1; x dx = du/2
No need to introduce a square-root branch for x.
(1/2)∫(u − 1)/u du
All factors are now in u.
(1/2)[u − ln u] + C
Here u = x² + 1 > 0.
(x² + 1)/2 − (1/2)ln(x² + 1) + C
Differentiation verifies the answer even at x = 0.
11 · A remaining quadratic factor
With u = x² + 2, rewrite ∫x³(x² + 2)² dx.
Hint
Use x³ dx = x²(x dx).
Worked solution
(1/2)∫(u − 2)u² du, giving u⁴/8 − u³/3 + C, then replace u with x² + 2.
12 · Avoid a false global inverse
Does u = x² + 1 imply x = √(u − 1) for every real x?
Hint
Consider negative x.
Worked solution
No. That chooses the nonnegative branch. In the worked example, using x² and x dx directly avoids this unjustified restriction.
09 / Choose a substitution that simplifies
Different substitutions can lead to equivalent answers.
A good candidate often removes a repeated bracket, a root or an exponential. Inspect what its derivative contributes. The choice must make the complete transformed expression simpler, not merely rename one awkward part.
Compare choices for ∫1/(2x + 6) dx.Worked example
u = 2x + 6 gives (1/2)ln|2x + 6| + C
Here dx = du/2.
v = x + 3 gives (1/2)ln|x + 3| + D
Here 2x + 6 = 2v and dx = dv.
The answers differ by (1/2)ln 2
That fixed difference is absorbed in the integration constant.
13 · Choose an effective substitution
What substitution simplifies ∫x/√(x² + 9) dx?
Hint
Try the expression under the root.
Worked solution
u = x² + 9 gives (1/2)∫u⁻¹ᐟ² du, so the result is √(x² + 9) + C.
10 / Check in the original variable
A derivative check catches missing factors and unfinished substitutions.
Differentiate the final expression in x. It should reproduce the original integrand, including every coefficient and sign. A mixture such as ∫x√u du is incomplete unless x has been explicitly expressed as a function of u.
Keep the domain in view: denominators must stay nonzero, roots must be real, and using an inverse substitution may require a branch restriction. Definite integrals need transformed endpoints too; the next lesson handles those separately.
14 · Repair a mixed-variable line
For ∫x√(x + 2) dx with u = x + 2, a student writes ∫x√u du. Complete the substitution.
Hint
Replace the remaining x.
Worked solution
∫(u − 2)√u du.
15 · Detect a missing differential
For ∫e²ˣ/(1 + eˣ) dx with u = eˣ, a student writes ∫u²/(1 + u) du. What is missing?
Hint
dx is not du.
Worked solution
dx = du/u, so the correct transformed integral is ∫u/(1 + u) du.
11 / Use a complete substitution ledger
Choose u, transform, simplify, integrate and return to x.
- State the substitution and its derivative.
- Rewrite every remaining factor and the differential.
- Simplify the integral in one variable.
- Integrate and substitute back.
- Check the derivative and valid domain.
16 · Finish a new substitution
Find ∫x√(x + 4) dx.
Hint
Use u = x + 4, then expand (u − 4)√u.
Worked solution
(2/5)(x + 4)⁵ᐟ² − (8/3)(x + 4)³ᐟ² + C, on x > −4.
Section 1 of 11 · Change the variable consistently