Integration by Parts: Sure Questions (Part 1)

This lesson sets up the integration by parts formula and the ILATE rule for choosing functions, then works through eight common Class 12 questions in order.

The formula and the ILATE rule

Integration by parts states

$\int u \, v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx,$

where $u$ is the first function and $v$ is the second. To decide which factor is the first function, use the ILATE order of preference: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. Whichever type comes earlier in this list is taken as the first function.

Question 1: $\int x^2 e^{x} \, dx$

Here the algebraic factor $x^2$ comes before the exponential $e^{x}$ in ILATE, so $u = x^2$ and $v = e^{x}$:

$\int x^2 e^{x} \, dx = x^2 e^{x} - \int 2x \, e^{x} \, dx = x^2 e^{x} - 2 \int x \, e^{x} \, dx.$

Apply parts again to $\int x \, e^{x} \, dx$ with $u = x$, $v = e^{x}$:

$\int x \, e^{x} \, dx = x e^{x} - \int e^{x} \, dx = x e^{x} - e^{x}.$

Substituting back,

$\int x^2 e^{x} \, dx = x^2 e^{x} - 2(x e^{x} - e^{x}) + C = e^{x}(x^2 - 2x + 2) + C.$

Question 2: $\int x \log x \, dx$

Logarithmic beats algebraic in ILATE, so $u = \log x$ and $v = x$:

$\int x \log x \, dx = \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx = \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx.$

Therefore

$\int x \log x \, dx = \frac{x^2}{2} \log x - \frac{x^2}{4} + C.$

Question 3: $\int x \log 2x \, dx$

Again take $u = \log 2x$ and $v = x$. Since $\dfrac{d}{dx}\log 2x = \dfrac{1}{2x} \cdot 2 = \dfrac{1}{x}$,

$\int x \log 2x \, dx = \log 2x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx = \frac{x^2}{2} \log 2x - \frac{1}{2} \int x \, dx.$

So

$\int x \log 2x \, dx = \frac{x^2}{2} \log 2x - \frac{x^2}{4} + C.$

Question 4: $\int x (\log x)^2 \, dx$

Take $u = (\log x)^2$ and $v = x$. Using $\dfrac{d}{dx}(\log x)^2 = \dfrac{2 \log x}{x}$,

$\int x (\log x)^2 \, dx = \frac{x^2}{2} (\log x)^2 - \int \frac{2 \log x}{x} \cdot \frac{x^2}{2} \, dx = \frac{x^2}{2} (\log x)^2 - \int x \log x \, dx.$

From Question 2, $\displaystyle\int x \log x \, dx = \frac{x^2}{2} \log x - \frac{x^2}{4}$, so

$\int x (\log x)^2 \, dx = \frac{x^2}{2} (\log x)^2 - \frac{x^2}{2} \log x + \frac{x^2}{4} + C.$

Taking $\dfrac{x^2}{4}$ as a common factor,

$\int x (\log x)^2 \, dx = \frac{x^2}{4} \left( 2 (\log x)^2 - 2 \log x + 1 \right) + C.$

Question 5: $\int \log |x| \, dx$

With a single logarithm, take the second function as $1$, so $u = \log |x|$ and $v = 1$:

$\int \log |x| \, dx = \log |x| \cdot x - \int \frac{1}{x} \cdot x \, dx = x \log |x| - \int 1 \, dx.$

Hence

$\int \log |x| \, dx = x \log |x| - x + C = x \left( \log |x| - 1 \right) + C.$

Question 6: $\int e^{\sqrt{x}} \, dx$

First substitute $\sqrt{x} = t$, so $x = t^2$ and $dx = 2t \, dt$:

$\int e^{\sqrt{x}} \, dx = \int e^{t} \cdot 2t \, dt = 2 \int t \, e^{t} \, dt.$

Now integrate by parts with $u = t$, $v = e^{t}$:

$2 \int t \, e^{t} \, dt = 2 \left( t e^{t} - \int e^{t} \, dt \right) = 2 e^{t}(t - 1) + C.$

Replacing $t = \sqrt{x}$,

$\int e^{\sqrt{x}} \, dx = 2 e^{\sqrt{x}} \left( \sqrt{x} - 1 \right) + C.$

Question 7: $\int \sqrt{x} \, e^{\sqrt{x}} \, dx$

Substitute $\sqrt{x} = t$ again, so $x = t^2$ and $dx = 2t \, dt$:

$\int \sqrt{x} \, e^{\sqrt{x}} \, dx = \int t \, e^{t} \cdot 2t \, dt = 2 \int t^2 e^{t} \, dt.$

Using the result from Question 1, $\displaystyle\int t^2 e^{t} \, dt = e^{t}(t^2 - 2t + 2)$, so

$2 \int t^2 e^{t} \, dt = 2 e^{t}(t^2 - 2t + 2) + C.$

Replacing $t = \sqrt{x}$ (so $t^2 = x$),

$\int \sqrt{x} \, e^{\sqrt{x}} \, dx = 2 e^{\sqrt{x}} \left( x - 2\sqrt{x} + 2 \right) + C.$

Question 8: if $\int |x| \, dx = k \, x |x| + C$, find $k$

Write $\int |x| \, dx = \int |x| \cdot 1 \, dx$ and integrate by parts with $u = |x|$, $v = 1$. Using $\dfrac{d}{dx}|x| = \dfrac{x}{|x|}$,

$\int |x| \, dx = |x| \cdot x - \int \frac{x}{|x|} \cdot x \, dx = x |x| - \int \frac{x^2}{|x|} \, dx.$

Since $x^2 = |x|^2$, the remaining integral is just $\int |x| \, dx$:

$\int |x| \, dx = x |x| - \int |x| \, dx + C_1.$

Transpose the integral to the left side:

$2 \int |x| \, dx = x |x| + C_1 \quad \Longrightarrow \quad \int |x| \, dx = \tfrac{1}{2} x |x| + C.$

Comparing with the given form $k \, x |x| + C$ shows $k = \tfrac{1}{2}$.

Key takeaways

• Choose the first function by the ILATE order: Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential.
• When one application of parts leaves another product, apply parts again, as with $\int x^2 e^{x} \, dx = e^{x}(x^2 - 2x + 2) + C$.
• For $e^{\sqrt{x}}$ and $\sqrt{x}\,e^{\sqrt{x}}$, substitute $\sqrt{x} = t$ first, then integrate by parts.
• If parts reproduces the original integral, move it to the other side and solve, giving $\int |x| \, dx = \tfrac{1}{2} x |x| + C$.