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NDA Maths · Indefinite Integration

Integration by Partial Fractions

Partial fractions break a single rational function into a sum of simpler fractions, each of which integrates to a logarithm or an arctangent.

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Why this matters

7 PYQs sit here, and the NDA reuses one shape relentlessly: 1/(x(xⁿ+1)). Beyond that, two variations recur — substitute a trig expression first and THEN decompose, and the 'express the numerator using the denominator and its derivative' trick. All three reduce to splitting, then integrating term by term.

Test yourself — a quick 13-question Indefinite Integration recall quiz, instant score.

Concept 1 of 4: Decomposition and the Cover-Up Method

A proper rational function whose denominator factors can be rewritten as a sum of fractions, one per factor. The cover-up method reads off each numerator instantly by plugging in the root that kills the other factors.

Definition

For distinct linear factors,

p(x)(x−a)(x−b)=Ax−a+Bx−b.\dfrac{p(x)}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}.
Cover-up: to get AA, cover the (x−a)(x-a) factor and evaluate the rest at x=ax=a; likewise for BB at x=bx=b. Each piece then integrates as ∫Ax−a dx=Aln⁡∣x−a∣\int \dfrac{A}{x-a}\,dx = A\ln|x-a|. An irreducible quadratic factor x2+1x^2+1 needs a numerator of the form Cx+DCx+D and integrates to a log plus an arctan.

Linear-factor decomposition

p(x)(x−a)(x−b)=Ax−a+Bx−b\dfrac{p(x)}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}

Worked example

Evaluate ∫dx(x−1)(x+2)\displaystyle\int \dfrac{dx}{(x-1)(x+2)}.

Decompose only a PROPER fraction

If the numerator's degree is ≥\geq the denominator's, do polynomial division FIRST, then decompose the remainder. Skipping this gives a wrong split.

Concept 2 of 4: The Recurring 1 over x times x-to-the-n-plus-1 Family

This exact shape appears again and again in NDA papers. The trick is not classical partial fractions — multiply top and bottom by x-to-the-(n-1) to manufacture the derivative of the denominator, turning it into two log terms.

Definition

For ∫dxx(xn+1)\displaystyle\int \dfrac{dx}{x(x^n+1)}, multiply numerator and denominator by xn−1x^{n-1}:

∫xn−1 dxxn(xn+1).\int \dfrac{x^{n-1}\,dx}{x^n(x^n+1)}.
Now 1xn(xn+1)=1xn−1xn+1\dfrac{1}{x^n(x^n+1)} = \dfrac{1}{x^n} - \dfrac{1}{x^n+1} (a one-line split), and xn−1 dx=1n d(xn)x^{n-1}\,dx = \tfrac{1}{n}\,d(x^n). With t=xnt=x^n it becomes 1n∫(1t−1t+1)dt\dfrac{1}{n}\int\big(\tfrac1t - \tfrac{1}{t+1}\big)dt, giving the standard answer
1nln⁡∣xnxn+1∣+C.\dfrac{1}{n}\ln\left|\dfrac{x^n}{x^n+1}\right| + C.

Closed form for the family

∫dxx(xn+1)=1nln⁡∣xnxn+1∣+C\int \dfrac{dx}{x(x^n+1)} = \dfrac{1}{n}\ln\left|\dfrac{x^n}{x^n+1}\right| + C

Worked example

Evaluate ∫dxx(x3+1)\displaystyle\int \dfrac{dx}{x(x^3+1)}.

The same idea in a real exam question:

NDA · Apr 2020 · Q88Moderate

Example 2 · Indefinite Integration · Integration by Partial Fractions

What is ∫dxx(xn+1)\int\frac{dx}{x(x^n+1)} equal to?

The 1 over n out front is easy to lose

The coefficient is 1n\dfrac{1}{n}, coming from xn−1dx=1nd(xn)x^{n-1}dx = \tfrac1n d(x^n). For n=7n=7 the answer carries 17\tfrac17; the no-coefficient option is the planted distractor.
Drill 2 more on the recurring 1 over x times x-to-the-n-plus-1 family

Concept 3 of 4: Substitute First, Then Decompose

A trig integrand can become a rational function in one substitution — then ordinary partial fractions finish it. The classic case is sine on top with a product of cosine-linear factors below: substitute u equals cosine.

Definition

For ∫sin⁡θ dθ(2+cos⁡θ)(3+4cos⁡θ)\displaystyle\int \dfrac{\sin\theta\,d\theta}{(2+\cos\theta)(3+4\cos\theta)}, put u=cos⁡θu=\cos\theta, du=−sin⁡θ dθdu = -\sin\theta\,d\theta. The integral becomes −∫du(2+u)(3+4u)-\displaystyle\int \dfrac{du}{(2+u)(3+4u)}, a rational function. Decompose 1(2+u)(3+4u)=A′2+u+B′3+4u\dfrac{1}{(2+u)(3+4u)} = \dfrac{A'}{2+u} + \dfrac{B'}{3+4u} by cover-up, integrate to logs, then put u=cos⁡θu=\cos\theta back. When the answer is written Aln⁡∣2+cos⁡θ∣+Bln⁡∣3+4cos⁡θ∣A\ln|2+\cos\theta| + B\ln|3+4\cos\theta|, the ln⁡∣3+4cos⁡θ∣\ln|3+4\cos\theta| coefficient picks up an extra 14\tfrac14 from the chain factor inside that factor.

Trig-to-rational substitution

∫sin⁡θ dθf(cos⁡θ)=−∫duf(u),u=cos⁡θ\int \dfrac{\sin\theta\,d\theta}{f(\cos\theta)} = -\int \dfrac{du}{f(u)},\quad u=\cos\theta

Worked example

Evaluate ∫sin⁡θ dθ(1+cos⁡θ)(2+cos⁡θ)\displaystyle\int \dfrac{\sin\theta\,d\theta}{(1+\cos\theta)(2+\cos\theta)}.

The same idea in a real exam question:

NDA · Apr 2026 · Q91Hard

Example 3 · Indefinite Integration · Integration by Partial Fractions

for the items that follow: Let ∫sin⁡θ dθ(2+cos⁡θ)(3+4cos⁡θ)=Aln⁡∣2+cos⁡θ∣+Bln⁡∣3+4cos⁡θ∣\int\dfrac{\sin\theta\,d\theta}{(2+\cos\theta)(3+4\cos\theta)}=A\ln|2+\cos\theta|+B\ln|3+4\cos\theta|.
What is the value of A?

Carry the minus from du, and the chain factor

u=cos⁡θu=\cos\theta gives du=−sin⁡θ dθdu=-\sin\theta\,d\theta — the leading minus stays. And a factor like 3+4cos⁡θ3+4\cos\theta contributes a 14\tfrac14 to its log coefficient because ddu(3+4u)=4\dfrac{d}{du}(3+4u)=4.
Drill 1 more on substitute first, then decompose

Concept 4 of 4: Express the Numerator via the Denominator and Its Derivative

When both numerator and denominator are linear combinations of sine and cosine, you cannot use cover-up. Instead write the numerator as A times the denominator plus B times the denominator's derivative — then the integral splits into an x-term and a log-term.

Definition

For ∫pcos⁡x+qsin⁡xacos⁡x+bsin⁡x dx\displaystyle\int \dfrac{p\cos x + q\sin x}{a\cos x + b\sin x}\,dx, set

pcos⁡x+qsin⁡x=A (acos⁡x+bsin⁡x)+B ddx(acos⁡x+bsin⁡x).p\cos x + q\sin x = A\,(a\cos x + b\sin x) + B\,\dfrac{d}{dx}(a\cos x + b\sin x).
Match the cos⁡x\cos x and sin⁡x\sin x coefficients to solve the 2×22\times 2 system for A,BA, B. Then
∫=A ⁣∫1 dx+B ⁣∫(denominator)′denominator dx=A x+Bln⁡∣acos⁡x+bsin⁡x∣+C,\int = A\!\int 1\,dx + B\!\int \dfrac{(\text{denominator})'}{\text{denominator}}\,dx = A\,x + B\ln|a\cos x + b\sin x| + C,
the second piece being the f′/ff'/f log pattern. The same "numerator = combination of the denominator" move works on SURD denominators: for ∫x2−1x2+1 dx\int\dfrac{x^2-1}{\sqrt{x^2+1}}\,dx, write x2−1=(x2+1)−2x^2-1=(x^2+1)-2, so the integrand is x2+1−2x2+1\sqrt{x^2+1}-\dfrac{2}{\sqrt{x^2+1}} — two standard forms, ∫x2+a2 dx=x2x2+a2+a22ln⁡∣x+x2+a2∣\int\sqrt{x^2+a^2}\,dx=\tfrac x2\sqrt{x^2+a^2}+\tfrac{a^2}{2}\ln\big|x+\sqrt{x^2+a^2}\big| and ∫dxx2+a2=ln⁡∣x+x2+a2∣\int\dfrac{dx}{\sqrt{x^2+a^2}}=\ln\big|x+\sqrt{x^2+a^2}\big|. In general N(x)=A⋅D(x)+(remainder)N(x)=A\cdot D(x)+(\text{remainder}) with DD the expression under the root.

Numerator as denom + derivative

N(x)=A D(x)+B D′(x) ⇒ ∫ND dx=Ax+Bln⁡∣D∣+CN(x) = A\,D(x) + B\,D'(x)\ \Rightarrow\ \int \dfrac{N}{D}\,dx = A x + B\ln|D| + C

Worked example

Express 5cos⁡x+sin⁡x5\cos x + \sin x as A(cos⁡x+sin⁡x)+Bddx(cos⁡x+sin⁡x)A(\cos x + \sin x) + B\dfrac{d}{dx}(\cos x + \sin x), and integrate 5cos⁡x+sin⁡xcos⁡x+sin⁡x\dfrac{5\cos x + \sin x}{\cos x + \sin x}.

The same idea in a real exam question:

NDA · Apr 2024 · Q81Moderate

Example 4 · Indefinite Integration · Integration by Partial Fractions

Given that ∫3cos⁡x+4sin⁡x2cos⁡x+5sin⁡x dx=αx29+β29ln⁡∣2cos⁡x+5sin⁡x∣+c\displaystyle\int\dfrac{3\cos x+4\sin x}{2\cos x+5\sin x}\,dx=\dfrac{\alpha x}{29}+\dfrac{\beta}{29}\ln|2\cos x+5\sin x|+c.
What is the value of α\alpha?

Watch the sign in the denominator's derivative

ddx(2cos⁡x+5sin⁡x)=−2sin⁡x+5cos⁡x\dfrac{d}{dx}(2\cos x + 5\sin x) = -2\sin x + 5\cos x — the cosine term's derivative is −sin⁡-\sin. A sign slip in the 2×22\times2 system swaps AA and BB, the exact distractor the paired items test.
Drill 1 more on express the numerator via the denominator and its derivative

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

  • Decomposition and the Cover-Up Method

    Linear-factor decomposition

    p(x)(x−a)(x−b)=Ax−a+Bx−b\dfrac{p(x)}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}
  • The Recurring 1 over x times x-to-the-n-plus-1 Family

    Closed form for the family

    ∫dxx(xn+1)=1nln⁡∣xnxn+1∣+C\int \dfrac{dx}{x(x^n+1)} = \dfrac{1}{n}\ln\left|\dfrac{x^n}{x^n+1}\right| + C
  • Substitute First, Then Decompose

    Trig-to-rational substitution

    ∫sin⁡θ dθf(cos⁡θ)=−∫duf(u),u=cos⁡θ\int \dfrac{\sin\theta\,d\theta}{f(\cos\theta)} = -\int \dfrac{du}{f(u)},\quad u=\cos\theta
  • Express the Numerator via the Denominator and Its Derivative

    Numerator as denom + derivative

    N(x)=A D(x)+B D′(x) ⇒ ∫ND dx=Ax+Bln⁡∣D∣+CN(x) = A\,D(x) + B\,D'(x)\ \Rightarrow\ \int \dfrac{N}{D}\,dx = A x + B\ln|D| + C

Watch out for (4)

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