A brief note on Methods of Integration . Read it before a test. Try each question at the end before you open its answer.
Inverse tangent: ∫ d x x 2 + a 2 = 1 a tan − 1 x a + C \int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C ∫ x 2 + a 2 d x = a 1 tan − 1 a x + C . Inverse sine: ∫ d x a 2 − x 2 = sin − 1 x a + C \int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\dfrac{x}{a} + C ∫ a 2 − x 2 d x = sin − 1 a x + C . Logarithms: ∫ d x x 2 − a 2 = 1 2 a ln ∣ x − a x + a ∣ + C \int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a}\ln\left|\dfrac{x - a}{x + a}\right| + C ∫ x 2 − a 2 d x = 2 a 1 ln x + a x − a + C and ∫ d x x 2 ± a 2 = ln ∣ x + x 2 ± a 2 ∣ + C \int \dfrac{dx}{\sqrt{x^2 \pm a^2}} = \ln\left|x + \sqrt{x^2 \pm a^2}\right| + C ∫ x 2 ± a 2 d x = ln x + x 2 ± a 2 + C . A quadratic: complete the square, then put u = x + b 2 a u = x + \dfrac{b}{2a} u = x + 2 a b . A linear top: write p x + q = A ( 2 a x + b ) + B px + q = A(2ax + b) + B p x + q = A ( 2 a x + b ) + B . The first part gives a logarithm. Partial fractions Forms: A x − a \dfrac{A}{x - a} x − a A for a linear factor, all powers for a repeated factor, and B x + C x 2 + p 2 \dfrac{Bx + C}{x^2 + p^2} x 2 + p 2 B x + C for a quadratic factor. Cover-up rule: hide a linear factor and put its root in the rest, to find its constant. Improper fractions: if the top has degree at least that of the bottom, divide first. Integrals: A ln ∣ x − a ∣ A\ln|x - a| A ln ∣ x − a ∣ and − B x − a -\dfrac{B}{x - a} − x − a B . Integration by parts Formula: ∫ u d v = u v − ∫ v d u \int u\,dv = uv - \int v\,du ∫ u d v = uv − ∫ v d u . Choice of u u u : ILATE, which is Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. Alone: ln x \ln x ln x and tan − 1 x \tan^{-1}x tan − 1 x use d v = d x dv = dx d v = d x . A cycle: for e x sin x e^x\sin x e x sin x , use parts twice and solve for the integral. A shortcut: ∫ e x [ f ( x ) + f ′ ( x ) ] d x = e x f ( x ) + C \int e^x\bigl[f(x) + f'(x)\bigr]dx = e^x f(x) + C ∫ e x [ f ( x ) + f ′ ( x ) ] d x = e x f ( x ) + C . A worked example Find ∫ x e x d x \displaystyle\int x\,e^{x}\,dx ∫ x e x d x .
Choose. Take u = x u = x u = x and d v = e x d x dv = e^x\,dx d v = e x d x , so d u = d x du = dx d u = d x and v = e x v = e^x v = e x . Apply the formula. x e x − ∫ e x d x = x e x − e x + C x\,e^x - \int e^x\,dx = x\,e^x - e^x + C x e x − ∫ e x d x = x e x − e x + C . Practice with new numbers
A program to try The program checks the split of 1 ( x − 1 ) ( x + 2 ) \dfrac{1}{(x - 1)(x + 2)} ( x − 1 ) ( x + 2 ) 1 into two fractions at four values of x x x .
Run it. Do the two columns agree?
1. Change the second term to a plus sign. Do they still agree?
Mistakes to avoid Wrong constant in front. 1 a \dfrac{1}{a} a 1 for the inverse tangent, and 1 2 a \dfrac{1}{2a} 2 a 1 for the logarithm. Not dividing first. An improper fraction cannot be split directly. A missing term. A repeated factor needs every power. A lost sign in parts. The formula is u v − ∫ v d u uv - \int v\,du uv − ∫ v d u . Check yourself Find ∫ d x x 2 + 36 \displaystyle\int \dfrac{dx}{x^2 + 36} ∫ x 2 + 36 d x . Stuck? Show a hint Show the answerSplit 5 ( x − 2 ) ( x + 3 ) \dfrac{5}{(x - 2)(x + 3)} ( x − 2 ) ( x + 3 ) 5 into partial fractions. Stuck? Show a hint Show the answerFind ∫ x cos x d x \displaystyle\int x\cos x\,dx ∫ x cos x d x . Stuck? Show a hint Show the answerFind ∫ e x ( cos x − sin x ) d x \displaystyle\int e^x\left(\cos x - \sin x\right)dx ∫ e x ( cos x − sin x ) d x . Stuck? Show a hint Show the answerWhich part comes first in ILATE, a logarithm or a power of x x x ? Stuck? Show a hint Show the answerSources OpenStax, Calculus Volume 2 , Chapter 3, sections 3.1 and 3.4. NCERT, Mathematics Part II for Class XII, Chapter 7, the sections on integrals of some particular functions, partial fractions and parts.