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Techniques of integration · Choose and check integration by parts

Choose parts when the remaining integral improves

Inspect the inner derivative before treating a product by parts. Choose what to differentiate and integrate, then check whether the new integral is simpler or returns to one you can solve for.

  1. Check for a reverse-chain match first

    A product alone does not decide the method. Simplify and compare a composite function with its inner derivative; a constant mismatch can be compensated.

    ddxex2=2xex2.\dfrac{d}{dx}e^{x^2}=2xe^{x^2}.

    ddxe2x=2e2x.\dfrac{d}{dx}e^{2x}=2e^{2x}.

    Reverse chainA matching derivative can avoid parts

    For xex2xe^{x^2}, the factor x matches the exponent’s derivative up to a constant. Substitution gives 12ex2+C\tfrac12e^{x^2}+C. Do not multiply by a missing function of x as if it were a constant adjustment.

    Parts candidateA leftover factor may suit differentiation

    For xe2xxe^{2x}, integrating the exponential alone leaves the factor x. Parts with u=xu=x and dv=e2x dxdv=e^{2x}\,dx reduces the polynomial degree. Other expressions may need algebra, identities or partial fractions before either method helps.

    These are useful structural clues, not an exhaustive or exclusive decision tree. Some integrals admit several valid methods, and a supplied substitution or preceding result can guide the intended route.

  2. Choose u and dv by the work they leave

    Choose u to differentiate and dv to integrate. Make sure v can be found and inspect the remaining integral before committing to repeated steps.

    ∫u dv=uv−∫v du.\int u\,dv=uv-\int v\,du.

    Useful choiceAim for simpler remaining work

    For ∫xex dx\int xe^x\,dx, choosing u=xu=x, dv=ex dxdv=e^x\,dx leaves ∫ex dx\int e^x\,dx. Reversing the roles leaves 12∫x2ex dx\tfrac12\int x^2e^x\,dx, which is harder. A valid use of the formula can still be an inefficient choice.

    LIATE guideUse the mnemonic with judgement

    Logarithmic, inverse trigonometric, algebraic, trigonometric and exponential functions give a useful preference for u. There are exceptions. Compare the resulting integral rather than treating the order as a compulsory rule.

    A standalone function can be written as a product with 1. For ln⁡x\ln x, choose u=ln⁡xu=\ln x and dv=dxdv=dx on x>0x>0; integration is not the same operation as differentiation.

  3. Repeat purposefully and check the result

    Keep the minus sign and coefficients. Repeated parts may lower a polynomial degree or bring back the original integral, which can then be collected algebraically.

    F′(x)=original integrand.F^{\prime}(x)=\text{original integrand}.

    Repeated partsTrack the integral you started with

    If it returns, name it and move its multiple to the same side before solving. Combine arbitrary constants consistently. Repeating the same unsuccessful choice without inspecting the new integral does not establish progress.

    VerificationRecover the complete original integrand

    Differentiate the final expression using product and chain rules. Carry the original domain restrictions, and include an arbitrary constant for an indefinite integral. A product is not integrated by multiplying its separate antiderivatives.

    The worked cases below concern indefinite integrals. Definite integrals also require consistent evaluation of the boundary term and the remaining integral on their stated interval.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

Example 17(c): make the factor 1 visible

For ∫ln⁡x dx\int\ln x\,dx on x>0x>0, decide which factor to differentiate after writing the integrand as (ln⁡x)×1(\ln x)\times1.

Show reasoning

  1. 1. Choose factors that simplify

    Differentiating ln x gives 1/x, while integrating 1 gives x. Those factors cancel in the remaining integral.

    u=ln⁡x,dv=dx,du=1x dx,v=x.\begin{aligned}u&=\ln x,&dv&=dx,\\du&=\frac1x\,dx,&v&=x.\end{aligned}

  2. 2. Apply parts and retain the domain

    The remaining integrand is 1. The logarithm and the original integral are defined here on x>0.

    ∫ln⁡x dx=xln⁡x−∫1 dx=xln⁡x−x+C.\begin{aligned}&\int\ln x\,dx\\&=x\ln x-\int1\,dx\\&=x\ln x-x+C.\end{aligned}

  3. 3. Differentiate instead of guessing

    The product rule recovers ln x. The expression 1/x is its derivative, not its antiderivative.

    ddx(xln⁡x−x)=ln⁡x+1−1=ln⁡x.\begin{aligned}&\frac{d}{dx}(x\ln x-x)\\&=\ln x+1-1=\ln x.\end{aligned}
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Example 17(e): solve the returning integral

For I=∫e2xsin⁡x dxI=\int e^{2x}\sin x\,dx, choose u=sin⁡xu=\sin x and dv=e2x dxdv=e^{2x}\,dx. Predict what will happen after applying parts a second time.

Show reasoning

  1. 1. Name the remaining integral

    The exponential integrates to e2x/2e^{2x}/2. Let J denote an antiderivative of e2xcos⁡xe^{2x}\cos x; retain the factor 1/2 and an arbitrary constant.

    J=∫e2xcos⁡x dx,I=12e2xsin⁡x−12J+C1.\begin{aligned}J&=\int e^{2x}\cos x\,dx,\\I&=\frac12e^{2x}\sin x\\&\quad-\frac12J+C_1.\end{aligned}

  2. 2. Apply parts again and keep the sign

    Choose u=cos⁡xu=\cos x for J. Its derivative is −sin⁡x-\sin x, so subtracting the remaining integral produces a positive multiple of I.

    J=12e2xcos⁡x+12I+C2.\begin{aligned}J&=\frac12e^{2x}\cos x\\&\quad+\frac12I+C_2.\end{aligned}

  3. 3. Collect the original integral

    Substitute for J, move −I/4 to the left and combine the arbitrary constants. The original integral has returned; it is now an algebraic equation rather than another instruction to repeat parts.

    54I=12e2xsin⁡x−14e2xcos⁡x+C3,I=25e2xsin⁡x−15e2xcos⁡x+C.\begin{aligned}\frac54I&=\frac12e^{2x}\sin x\\&\quad-\frac14e^{2x}\cos x+C_3,\\I&=\frac25e^{2x}\sin x\\&\quad-\frac15e^{2x}\cos x+C.\end{aligned}

  4. 4. Check both product-rule terms

    Let F be the final antiderivative without C. Its cosine terms cancel and the sine coefficients add to 1, recovering the original integrand for every real x.

    F′(x)=45e2xsin⁡x+25e2xcos⁡x−25e2xcos⁡x+15e2xsin⁡x=e2xsin⁡x.\begin{aligned}F^{\prime}(x)&=\frac45e^{2x}\sin x\\&\quad+\frac25e^{2x}\cos x\\&\quad-\frac25e^{2x}\cos x\\&\quad+\frac15e^{2x}\sin x\\&=e^{2x}\sin x.\end{aligned}
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