Choose and check integration by parts

Parts reverses the product rule

Choose the differentiated factor so that the remaining integral becomes simpler. Compute its derivative and the other factor’s antiderivative separately, then apply parts. Inspect the new integral before repeating: some problems return to the original integral, which can then be solved algebraically.

The product rule is $d(uv)/dx=u\,dv/dx+v\,du/dx$. Integrate and rearrange to obtain $\int u\,dv/dx\,dx=uv-\int v\,du/dx\,dx$. Choose $u$ to differentiate and $dv/dx$ to integrate. The source highlights inverse-trigonometric functions, logarithms and products of mixed functions. A standalone function can be multiplied by 1.

Compare the source two choices for $\int xe^x\,dx$. With $u=x$, $dv/dx=e^x$, $du/dx=1$, $v=e^x$, parts gives $xe^x-\int e^x\,dx=xe^x-e^x+C$. With $u=e^x$, $dv/dx=x$, $du/dx=e^x$, $v=x^2/2$, it gives $x^2e^x/2-\frac12\int x^2e^x\,dx$, which is harder. Choose the route that simplifies the remaining integral, sometimes by trial and error.

The decreasing preference LIATE is Logarithmic, Inverse trigonometric, Algebraic (including constants), Trigonometric, Exponential. The Teacher warns this is a guide with exceptions, not an absolute rule. Margin examples include $\ln(2x-1)$ and $\log_5x$, $\tan^{-1}x$ and $\sin^{-1}(2x)$, a polynomial and $4=4x^0$, trigonometric functions, $e^x$ and $2^x$. The Teacher crosses out confusing an inverse-sine integral with its derivative; multiplying the integrand by 1 makes parts available, as in Example 17(c).

The printed remarks refer to Example 16(c), Example 16(d,e) and then Example 17 after Example 18. In the current numbering these demonstrations are Example 17(c), Example 17(d,e), and Example 18 respectively. Their teaching points are preserved: introduce 1, repeat parts, and use a preceding derivative. Native numbering is retained in the source transcription.

A method checklist from the closing Teacher notes

For substitution, follow the four steps in Example 16(a): differentiate the relation; replace the integrand and differential completely; integrate in the new variable; return to the original variable. In Example 16(b), implicit differentiation of $u^2=x-2$ makes $dx/du=2u$ clear; use long division after substitution. A triangle recovers trigonometric ratios, with signed identities covering negative inputs. The two square-root substitutions in Example 16 use the stated angle intervals to control the radical sign.

The Teacher mastery notes add long division for improper fractions and rewriting a denominator as a power: $\int x/\sqrt{x^2-1}\,dx=\frac12\int2x(x^2-1)^{-1/2}\,dx=\sqrt{x^2-1}+C$ on $|x|>1$. The final route-map annotations reinforce that $\int\ln x\,dx\ne1/x+C$; instead parts gives $x\ln x-x+C$ for $x>0$. They contrast the reverse-chain integral $\frac12\int2xe^{x^2}\,dx=\frac12e^{x^2}+C$ with the product $\int xe^x\,dx$, which uses parts as above. Self-Practice 9 answers the earlier motivating radical integral in both sine and cosine substitutions.

Choose from the inner derivative, not the appearance of a product

Compare $\int xe^{x^2}\,dx$ and $\int xe^{2x}\,dx$. Choose an efficient method for each, explain how the inner derivative affects the choice, and verify both answers by differentiation.

After your attempt: compare the reasoning

For the first integral, the derivative of the exponent is $2x$, already present up to a constant. Substitution $u=x^2$ gives $\tfrac12e^{x^2}+C$.

For the second, the derivative of $2x$ is constant, so the factor x remains. Parts with $u=x$, $dv=e^{2x}\,dx$ gives

$\frac{x}{2}e^{2x}-\frac14e^{2x}+C.$

Differentiating the two answers gives $xe^{x^2}$ and $\tfrac12e^{2x}+(x-\tfrac12)e^{2x}=xe^{2x}$ respectively. Both integrands look like products; the useful distinction is their structure under the chain and product rules.