The chain rule differentiates composite functions: the derivative of sin(x2) is cos(x2)⋅2x — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given cos(x2)⋅2x, find the original function. That's what u substitution does — it reverses the chain rule.
The Core Intuition
When an integral looks like "a function times the derivative of its inside," substitute the inside with u and the derivative of the inside with du. Consider:
∫2xcos(x2)dx
Here 2x is the derivative of x2, and x2 is the inside of cos(x2). Let u=x2, so du=2xdx:
∫cos(u)du=sin(u)+C=sin(x2)+C
Check: the derivative of sin(x2) is cos(x2)⋅2x.
The Precise Statement
∫f(g(x))⋅g′(x)dx=∫f(u)duwhere u=g(x),du=g′(x)dx
Valid provided g is differentiable and the resulting integral in u is simpler.
The Step-by-Step Method
Identify a function g(x) whose derivative g′(x) also appears (possibly up to a constant factor).
Setu=g(x), compute du=g′(x)dx.
Rewrite the entire integral in u and du — every x and dx must be replaced.
Integrate with respect to u.
Substitute backu=g(x).
Watch out
You cannot mix variables. If any x remains after substitution, you chose the wrong u (or must solve for x in terms of u — rare).
A Second Example (with a constant factor)
Evaluate ∫xx2+1dx. Let u=x2+1, so xdx=21du:
∫u⋅21du=21⋅32u3/2+C=31(x2+1)3/2+C
When Does It Work?
When the integrand is something times the derivative of something inside. Common patterns:
x⋅f(x2) — derivative of x2 is 2x, so u=x2
eg(x)⋅g′(x) — derivative of g(x) appears
g(x)g′(x) — leads to log∣g(x)∣
Tip
If stuck, differentiate a candidate "inside" function in your head. If its derivative (up to a constant) appears, that's your u.
The Definite Integral Case
Either change the limits (when x=a, u=g(a); when x=b, u=g(b); then integrate in u), or integrate in u, substitute back, and use the original limits. Changing limits is cleaner:
Don't confuse du with Δu. du is a differential — the exact relationship du=g′(x)dx that holds inside the integral. Treat it algebraically: multiply, divide, and substitute freely.
U-substitution, taught in the CBSE Class 12 Integrals chapter as the method of substitution, is one of the very first integration techniques students learn after the standard formulas, and "integration by substitution class 12 examples" is a heavily searched revision topic. It remains equally essential for solving integral calculus problems in JEE Main and JEE Advanced.
The key idea is U Substitution — noticing that the derivative of logx appears in the denominator.
Let u=logx. Then du=x1dx. The denominator x+xlogx=x(1+logx)=x(1+u).
The integral becomes:
∫x(1+u)1dx=∫1+u1du
Integrating:
∫1+u1du=log∣1+u∣+C
Substitute back u=logx:
log∣1+logx∣+C
✓Final answer
The integral is log∣1+logx∣+C.
The key idea is to factor x from the denominator and then use the substitution u=1+logx, which simplifies the integral to ∫udu=log∣u∣+C. The final result is log∣1+logx∣+C.
We start with the integral:
∫x+xlogx1dx
The denominator has a common factor of x in both terms. Factor it out:
∫x(1+logx)1dx
Now, why would we think of substitution here? The expression 1+logx appears inside the denominator, and its derivative is x1, which is also present in the integrand. This is the classic signal for a u-substitution: when you see a function and its derivative (up to a constant factor) multiplied together.
Let u=1+logx. Then differentiate:
dxdu=x1⇒du=x1dx
The integral becomes:
∫x(1+logx)1dx=∫u1du
This is a standard integral:
∫u1du=log∣u∣+C
Now substitute back u=1+logx:
log∣1+logx∣+C
Watch out
A common mistake is to forget the absolute value in the logarithm. Since logx is defined only for x>0, and 1+logx could be negative for 0<x<e−1, the absolute value is necessary for the general antiderivative. However, if the domain is restricted to x>e−1, you can drop the absolute value.
Tip
Notice that we didn't need to expand or simplify anything beyond factoring. The substitution u=1+logx works because the derivative of logx is 1/x, which cancels the x in the denominator perfectly. This is a textbook example of the "function-derivative" pattern.
✓Final answer
The integral evaluates to log∣1+logx∣+C.
Method: Factor the denominator first, then substitute
Use this when a denominator can be factored to expose a g(x) whose derivative appears — here x+xlogx=x(1+logx).
Steps
Step 1: Factor to reveal the hidden structure.
x+xlogx1=x(1+logx)1=x1⋅1+logx1.
Step 2: Substitute the bracket.
Let u=1+logx; then du=x1dx, exactly the leftover x1dx. The integral becomes ∫udu.
Step 3: Integrate to a logarithm and back-substitute.
∫udu=log∣u∣+C⇒log∣1+logx∣+C.
Factoring is the move students miss — without it the x1dx pattern stays hidden.
Common Mistakes
Mistake 1: Not factoring the denominator.
Why it's wrong: x+xlogx looks unfamiliar, but factoring to x(1+logx) reveals the x1 needed for substitution. Correct approach: always try to factor before deciding an integral is hard.
Mistake 2: Substituting u=logx instead of u=1+logx.
Why it's wrong: after factoring you have 1+logx1, so the cleaner substitution is u=1+logx (also giving du=x1dx). Correct approach: let u be the full bracket in the denominator.
Mistake 3: Forgetting the absolute value in the log.