Leibniz (Fraction) Notation Examples

Find .

The coefficient 4 is the derivative of the exponent 4x:

4e^4x.

Let u = 4x. Then the derivative of u with respect to x is 4. In symbols,

Thinking of  as a fraction and multiplying both sides by dx, we can write

du = 4dx.

Now we can rewrite the integral using u instead of x.

We know that

Putting the original variable back in, we have

e^4x + C.

We conclude that

Find .

Let

u = 5x².

Then

and

du = 10xdx.

We substitute variables, integrate, then put the original variable back:

For the integral, (a) identify u and du and (b) integrate by substitution.

(a) We want u to be the most complicated thing that is still an "inside" function, so let

u = 5x.

Then

and

du = 5dx.

(b) We change variables:

We integrate:

And finally we put the original variable back in:

= -cos(5x) + C.

We conclude

For the integral, (a) identify u and du and (b) integrate by substitution.

(a) The most complicated "inside" function is (3x + 2), so let

u = (3x + 2).

Then

so

du = 3dx.

(b) To change variables, we break up the coefficient 15 into 5 ⋅ 3 and move the 3 next to the dx so we can see where du is:

We undo the power rule to find the integral:

and put (3x + 2) back in for u:

u⁵ + C = (3x + 2)⁵ + C

Our final answer is

For the integral, (a) identify u and du and (b) integrate by substitution.

(a) The most complicated "inside" function is the exponent of e:

We let

u = 4x²

du = 8xdx

(b) We change variables, integrate using an appropriate pattern, and put the original variable back in:

Find .

Let

u = cos(2x)

du = -2sin(2x) dx

If we factor -4 into 2(-2), we can see how to do the substitution.

Find .

Take

u = 2x

du = 2dx

and move the factor  outside the integral sign after doing the substitution.

Things will work out nicely:

We write + C instead of + C because either expression will get us the same family of functions, so we may as well use the simpler one.

If you don't like the "multiplying by 1" trick, there's another option. Get an expression for dx that uses du, and substitute that expression for dx.

Example. Find without using the "multiplying by 1" trick.

Take

.

Now we can plug everything in and evaluate the integral:

We can use whichever trick you like better: multiply the integrand by a clever form of 1, or substitute an expression involving du for dx. As long as we do it right, we'll get the same answer either way.