Integration by Parts — Theory and Solved Exercises
Theoretical Recall
Integration by parts is obtained directly from the product rule for derivatives.
For two differentiable functions f and g:
\[(fg)'=f'g+fg'.\]Integrating both sides gives:
\[\int (fg)'\,dx = \int f'g\,dx + \int fg'\,dx.\]Therefore:
\[fg = \int f'g\,dx + \int fg'\,dx.\]Rearranging:
\[\int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx.\]This is the integration by parts formula.
A common alternative notation is:
\[\int u\,dv = uv-\int v\,du.\]Choosing the Functions
The goal is to choose the factor to differentiate so that it becomes simpler, while the other factor can be integrated easily.
A useful heuristic is the LIATE rule:
- L — Logarithmic functions
- I — Inverse trigonometric functions
- A — Algebraic functions
- T — Trigonometric functions
- E — Exponential functions
Functions appearing earlier in this list are often good candidates for the factor to differentiate.
This is a guideline rather than a theorem: the best choice is always the one that simplifies the resulting integral.
Author’s note: Integration by parts should not be viewed merely as a formula to memorize. Its real purpose is to transform an integral into another one whose structure is easier to handle.
Exercises
Exercise 1
Evaluate:
\[\int xe^x\,dx.\]Solution.
Choose:
f = x
and:
g′ = eˣ.
Then:
f′ = 1
and:
g = eˣ.
Using integration by parts:
\[\int xe^x\,dx = xe^x-\int e^x\,dx.\]Therefore:
\[\int xe^x\,dx = xe^x-e^x+C.\]Factor out the exponential:
\[\int xe^x\,dx = (x-1)e^x+C.\]Final Result
\[(x-1)e^x+C\]Exercise 2
Evaluate:
\[\int x\cos x\,dx.\]Solution.
Choose:
f = x
and:
g′ = cos x.
Then:
f′ = 1
and:
g = sin x.
Therefore:
\[\int x\cos x\,dx = x\sin x-\int\sin x\,dx.\]Since:
\[\int\sin x\,dx=-\cos x,\]we obtain:
\[\int x\cos x\,dx = x\sin x+\cos x+C.\]Final Result
\[x\sin x+\cos x+C\]Exercise 3
Evaluate:
\[\int x\sin x\,dx.\]Solution.
Choose:
f = x
and:
g′ = sin x.
Then:
f′ = 1
and:
g = −cos x.
Applying integration by parts:
\[\int x\sin x\,dx = -x\cos x+\int\cos x\,dx.\]Therefore:
\[\int x\sin x\,dx = -x\cos x+\sin x+C.\]Final Result
\[-x\cos x+\sin x+C\]Exercise 4
Evaluate:
\[\int xe^{2x}\,dx.\]Solution.
Choose:
f = x
and:
g′ = e²ˣ.
Then:
f′ = 1.
Moreover:
\[g = \int e^{2x}\,dx = \frac12e^{2x}.\]Integration by parts gives:
\[\int xe^{2x}\,dx = \frac{x}{2}e^{2x} - \frac12\int e^{2x}\,dx.\]Hence:
\[\int xe^{2x}\,dx = \frac{x}{2}e^{2x} - \frac14e^{2x} + C.\]Factoring:
\[\int xe^{2x}\,dx = \left( \frac{x}{2} - \frac14 \right)e^{2x} + C.\]Final Result
\[\left(\frac{x}{2}-\frac14\right)e^{2x}+C\]Exercise 5
Evaluate:
\[\int\ln x\,dx.\]Solution.
Although the integrand appears to contain only one function, write:
\[\ln x = (\ln x)\cdot1.\]Choose:
f = ln x
and:
g′ = 1.
Then:
\[f' = \frac1x,\]and:
g = x.
Integration by parts gives:
\[\int\ln x\,dx = x\ln x - \int x\frac1x\,dx.\]Thus:
\[\int\ln x\,dx = x\ln x-\int1\,dx.\]Therefore:
\[\int\ln x\,dx = x\ln x-x+C.\]Final Result
\[x\ln x-x+C\]Exercise 6
Evaluate:
\[\int x^2e^x\,dx.\]Solution.
Choose:
f = x²
and:
g′ = eˣ.
Then:
f′ = 2x
and:
g = eˣ.
Therefore:
\[\int x^2e^x\,dx = x^2e^x - 2\int xe^x\,dx.\]The remaining integral still requires integration by parts.
From Exercise 1:
\[\int xe^x\,dx = xe^x-e^x.\]Substituting:
\[\int x^2e^x\,dx = x^2e^x - 2(xe^x-e^x) + C.\]Expanding:
\[\int x^2e^x\,dx = x^2e^x - 2xe^x + 2e^x + C.\]Factoring out eˣ:
\[\int x^2e^x\,dx = (x^2-2x+2)e^x+C.\]Final Result
\[(x^2-2x+2)e^x+C\]Exercise 7
Evaluate:
\[\int x\ln x\,dx.\]Solution.
Choose the logarithmic factor for differentiation:
f = ln x.
Then choose:
g′ = x.
Therefore:
\[f' = \frac1x,\]and:
\[g = \frac{x^2}{2}.\]Integration by parts gives:
\[\int x\ln x\,dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2} \frac1x \,dx.\]Simplifying:
\[\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac12\int x\,dx.\]Therefore:
\[\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C.\]Final Result
\[\frac{x^2}{2}\ln x-\frac{x^2}{4}+C\]Exercise 8
Evaluate:
\[\int e^x\cos x\,dx.\]Solution.
Let:
\[I = \int e^x\cos x\,dx.\]Choose:
f = cos x
and:
g′ = eˣ.
Then:
f′ = −sin x
and:
g = eˣ.
Integration by parts gives:
\[I = e^x\cos x + \int e^x\sin x\,dx.\]Now define:
\[J = \int e^x\sin x\,dx.\]Apply integration by parts again, choosing:
f = sin x
and:
g′ = eˣ.
Then:
f′ = cos x
and:
g = eˣ.
Therefore:
\[J = e^x\sin x - \int e^x\cos x\,dx.\]Since the remaining integral is I:
\[J = e^x\sin x-I.\]Substitute this into the first equation:
\[I = e^x\cos x + e^x\sin x - I.\]Hence:
\[2I = e^x(\cos x+\sin x).\]Therefore:
\[I = \frac12e^x(\sin x+\cos x)+C.\]Final Result
\[\frac12e^x(\sin x+\cos x)+C\]Exercise 9
Evaluate:
\[\int e^x\sin x\,dx.\]Solution.
Let:
\[I = \int e^x\sin x\,dx.\]Choose:
f = sin x
and:
g′ = eˣ.
Then:
f′ = cos x
and:
g = eˣ.
Integration by parts gives:
\[I = e^x\sin x - \int e^x\cos x\,dx.\]Now apply integration by parts to the remaining integral:
\[\int e^x\cos x\,dx.\]Choose:
f = cos x
and:
g′ = eˣ.
Then:
\[\int e^x\cos x\,dx = e^x\cos x + \int e^x\sin x\,dx.\]The last integral is I, so:
\[\int e^x\cos x\,dx = e^x\cos x+I.\]Substitute into the original equation:
\[I = e^x\sin x - e^x\cos x - I.\]Therefore:
\[2I = e^x(\sin x-\cos x).\]Hence:
\[I = \frac12e^x(\sin x-\cos x)+C.\]Final Result
\[\frac12e^x(\sin x-\cos x)+C\]Exercise 10
Evaluate:
\[\int x^3e^x\,dx.\]Solution.
Let:
\[I = \int x^3e^x\,dx.\]Choose:
f = x³
and:
g′ = eˣ.
Then:
f′ = 3x²
and:
g = eˣ.
Therefore:
\[I = x^3e^x - 3\int x^2e^x\,dx.\]Now apply integration by parts to:
\[\int x^2e^x\,dx.\]We obtain:
\[\int x^2e^x\,dx = x^2e^x - 2\int xe^x\,dx.\]For the remaining integral:
\[\int xe^x\,dx = xe^x-e^x.\]Therefore:
\[\int x^2e^x\,dx = x^2e^x - 2xe^x + 2e^x.\]Substitute this expression into I:
\[I = x^3e^x - 3(x^2e^x-2xe^x+2e^x) + C.\]Expanding:
\[I = x^3e^x - 3x^2e^x + 6xe^x - 6e^x + C.\]Factor out eˣ:
\[I = (x^3-3x^2+6x-6)e^x+C.\]Final Result
\[(x^3-3x^2+6x-6)e^x+C\]