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Rational functions

Repeated quadratics and mixed factors

Material by the IntegralPaso project. Conditions and limits are stated in each guide; external teaching review is pending.

∑k=1mAkx+Bk(ax2+bx+c)k\sum_{k=1}^m\frac{A_kx+B_k}{(ax^2+bx+c)^k}

For a repeated quadratic include a linear numerator at every power. Combine templates when linear and quadratic factors occur together. Higher quadratic powers may need trigonometric substitution or reduction formulas.

Common mistake

Include all factors and powers even when some coefficients later turn out to be zero.

DEVELOPED GUIDE · 10 min

Solve repeated irreducible quadratic factors and understand why their numerators are linear.

When to choose this method

  • A quadratic factor has no real roots: its discriminant is negative.
  • For Q(x)², the general template includes (Ax+B)/Q and (Cx+D)/Q². Some coefficients can be zero.

Before calculating

  1. Look for the quadratic derivative first. Separate the remaining part.
  2. Use arctangent, reduction or trigonometric substitution for the remaining term.

A complete example, step by step

∫2x+1(x2+1)2 dx\int\frac{2x+1}{(x^2+1)^2}\,dx

  1. The quadratic x²+1 never vanishes over the reals and is squared.

    Δ=02−4⋅1⋅1=−4<0\Delta=0^2-4\cdot1\cdot1=-4<0

  2. Separate 2x, the derivative of x²+1, from the constant term.

    I=∫2x(x2+1)2 dx+∫dx(x2+1)2I=\int\frac{2x}{(x^2+1)^2}\,dx+\int\frac{dx}{(x^2+1)^2}

  3. For the first integral use u=x²+1 and du=2x dx.

    ∫u−2 du=−u−1=−1x2+1\int u^{-2}\,du=-u^{-1}=-\frac1{x^2+1}

  4. For the second use x=tan t on −π/2<t<π/2. Transform the differential too.

    dx=sec⁡2t dt,(1+x2)2=sec⁡4tdx=\sec^2t\,dt,\quad (1+x^2)^2=\sec^4t

  5. The quotient becomes cos²t; the double-angle identity allows integration.

    ∫cos⁡2t dt=t2+sin⁡(2t)4\int\cos^2t\,dt=\frac t2+\frac{\sin(2t)}4

  6. Return using t=arctan x and sin(2t)=2x/(1+x²).

    ∫dx(1+x2)2=12arctan⁡x+x2(1+x2)\int\frac{dx}{(1+x^2)^2}=\frac12\arctan x+\frac{x}{2(1+x^2)}

  7. Combine both parts; the formula is real for all x.

    I=12arctan⁡x+x−22(1+x2)+CI=\frac12\arctan x+\frac{x-2}{2(1+x^2)}+C

Check the result and domain

Check both parts separately. Differentiate the rational term with the quotient rule.

ddx[12arctan⁡x+x−22(1+x2)]=2x+1(1+x2)2\frac{d}{dx}\left[\frac12\arctan x+\frac{x-2}{2(1+x^2)}\right]=\frac{2x+1}{(1+x^2)^2}
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Now try it yourself

Solve on paper before opening the hint or answer. These are self-assessment activities; to check a typed answer, open Learn.

EXERCISE 1∫2x(1+x2)2 dx\int\frac{2x}{(1+x^2)^2}\,dx
Show a hint

Use u=1+x².

Check my result−11+x2+C-\frac1{1+x^2}+C
EXERCISE 2∫dx(1+x2)2\int\frac{dx}{(1+x^2)^2}
Show a hint

Repeat x=tan t.

Check my result12arctan⁡x+x2(1+x2)+C\frac12\arctan x+\frac{x}{2(1+x^2)}+C
EXERCISE 3∫x(1+x2)3 dx\int\frac{x}{(1+x^2)^3}\,dx
Show a hint

du=2x dx; compensate for the factor 2.

Check my result−14(1+x2)2+C-\frac1{4(1+x^2)^2}+C
Continue with interactive practice

Connect with other methods

Irreducible quadratic factors →Partial fractions →Trigonometric substitution: three radical forms →
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