3.5 Integration

Use substitution to expose an inner derivative, integration by parts to reduce a product, and partial fractions to split a rational function into standard terms.

Inspect the integrand before starting: ask what becomes simpler after differentiating one factor or replacing a repeated expression. Change limits and differentials consistently.

For ∫x ln x dx, take u=ln x and dv=x dx; for ∫2x/(1+x²)dx, use u=1+x².

A method is not justified because it produces more terms; the check is whether the new integral is genuinely simpler.

Rewrite products or powers of sin and cos using identities, then integrate each term. For an inner multiple angle, include the reciprocal derivative factor.

For even powers use half-angle identities; for odd powers save one factor and substitute. Keep absolute values in logarithmic results.

∫cos²x dx=x/2+sin2x/4+C; ∫sin(4x)dx=−cos(4x)/4+C.

A power on sin or cos is not an instruction to raise its antiderivative to a power.

After factoring the denominator, decompose a proper rational function into linear terms and irreducible quadratic terms. Integrate each using logarithm or arctangent forms.

Complete the square in a quadratic denominator before integrating. Repeated factors need one numerator for each power.

∫1/(x²+1)dx=tan⁻¹x+C, while ∫(2x)/(x²+1)dx=ln(x²+1)+C.

A quadratic denominator does not always produce a logarithm; if its numerator is constant after completing the square, an inverse tangent may appear.

For ∫x/(ax²+bx+c)dx, write the numerator as a multiple of the denominator derivative plus a constant remainder. The first part gives a logarithm; the remainder may require completing the square.

Differentiate the denominator first, solve a simple linear identity for the numerator split, and keep the denominator positive conditions for logs.

For ∫x/(x²+1)dx, x is half of (x²+1)′, so the result is ½ln(x²+1)+C.

The numerator need not equal the denominator derivative exactly; it may be a scaled derivative plus a leftover constant.

Use ∫du/(u²+a²)=(1/a)tan⁻¹(u/a)+C after rewriting the denominator as a square plus a positive constant.

Complete the square, factor the constant scale correctly, and use a substitution if the numerator contains the derivative of the shifted variable.

∫dx/[(x−2)²+9]=⅓tan⁻¹((x−2)/3)+C.

The coefficient is 1/a, not a; omitting it gives a derivative three times too large in the example.

Set u=g(x) when g′(x) appears, replace dx and every occurrence of x, integrate in u, then substitute back. A definite integral can instead use transformed limits.

Do not mix x and u in the same unfinished integral. Choose a substitution that removes the inner function rather than merely renaming it.

∫x√(x²+1)dx with u=x²+1 gives ½∫u^{1/2}du=(1/3)(x²+1)^{3/2}+C.

Substitution is incomplete if the differential factor is missing; the presence of g(x) alone is not enough.