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Integrals - Integration of a variety of functions by substitution, by partial fractions and by parts

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Integration by Substitution: This method simplifies an integral by changing the variable. If a function is of the form ∫f(g(x))g′(x)dx\int f(g(x)) g'(x) dx, we substitute t=g(x)t = g(x), making the integral ∫f(t)dt\int f(t) dt. Graphically, this corresponds to a transformation of the area under the curve to a simpler coordinate system.

Graph showing the area under a curve representing an integral.
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Integration by Partial Fractions: Rational functions P(x)Q(x)\frac{P(x)}{Q(x)} where the degree of P(x)P(x) is less than Q(x)Q(x) can be decomposed into simpler fractions. For example, 1(x−a)(x−b)=Ax−a+Bx−b\frac{1}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}. This splits a complex rational function into basic logarithmic integrals.

Flowchart showing decomposition of rational functions into partial fractions.
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Integration by Parts: Based on the product rule of differentiation, ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du. The ILATE rule (Inverse Trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential) is used to prioritize the choice of the function uu.

Visual representation of the components in integration by parts.
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Special Integrals: These are standard results for functions involving square roots and quadratic denominators. Remembering forms like ∫1x2+a2dx\int \frac{1}{x^2+a^2} dx is essential for solving problems involving completing the square.

Graph of the bell-shaped curve representing a standard rational integral form.

📐Formulae

∫dxx2−a2=12alog⁡∣x−ax+a∣+C\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C

∫dxa2−x2=12alog⁡∣a+xa−x∣+C\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C

∫dxx2+a2=1atan⁡−1(xa)+C\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C

∫dxa2−x2=sin⁡−1(xa)+C\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left( \frac{x}{a} \right) + C

∫dxx2−a2=log⁡∣x+x2−a2∣+C\int \frac{dx}{\sqrt{x^2 - a^2}} = \log \left| x + \sqrt{x^2 - a^2} \right| + C

∫dxx2+a2=log⁡∣x+x2+a2∣+C\int \frac{dx}{\sqrt{x^2 + a^2}} = \log \left| x + \sqrt{x^2 + a^2} \right| + C

∫u⋅v dx=u∫v dx−∫(dudx⋅∫v dx)dx\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \cdot \int v \, dx \right) dx

∫ex(f(x)+f′(x))dx=exf(x)+C\int e^x (f(x) + f'(x)) dx = e^x f(x) + C

💡Examples

Problem 1:

Evaluate ∫2x1+x2dx\int \frac{2x}{1 + x^2} dx using the method of substitution.

Solution:

Step 1: Let t=1+x2t = 1 + x^2. Step 2: Differentiate both sides with respect to xx: dtdx=2x\frac{dt}{dx} = 2x, which gives dt=2x dxdt = 2x \, dx. Step 3: Substitute tt and dtdt into the integral: ∫1tdt\int \frac{1}{t} dt. Step 4: Integrate the function: log⁡∣t∣+C\log |t| + C. Step 5: Substitute the value of tt back in terms of xx: log⁡∣1+x2∣+C\log |1 + x^2| + C.

Explanation:

This problem uses substitution because the derivative of the denominator (2x2x) is present in the numerator. By changing the variable, we transform a rational function into a standard reciprocal integral.

Problem 2:

Evaluate ∫xcos⁡x dx\int x \cos x \, dx using integration by parts.

Solution:

Step 1: Identify uu and vv using ILATE. Let u=xu = x (Algebraic) and dv=cos⁡x dxdv = \cos x \, dx (Trigonometric). Step 2: Calculate du=dxdu = dx and v=∫cos⁡x dx=sin⁡xv = \int \cos x \, dx = \sin x. Step 3: Apply the integration by parts formula: ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du. Step 4: Substitute the values: xsin⁡x−∫sin⁡x dxx \sin x - \int \sin x \, dx. Step 5: Integrate the remaining term: xsin⁡x−(−cos⁡x)+Cx \sin x - (-\cos x) + C. Step 6: Simplify: xsin⁡x+cos⁡x+Cx \sin x + \cos x + C.

Explanation:

Integration by parts is applied here because the integrand is a product of an algebraic function and a trigonometric function. Choosing u=xu=x allows the degree of the algebraic part to reduce to 1, making the second integral straightforward.

Problem 3:

Evaluate ∫1x2−16dx\int \frac{1}{x^2 - 16} dx using partial fractions.

Graph of the function 1/(x^2-16) showing vertical asymptotes at the roots of the denominator.

Solution:

  1. Factor the denominator: x2−16=(x−4)(x+4)x^2 - 16 = (x-4)(x+4).
  2. Decompose into partial fractions: 1(x−4)(x+4)=Ax−4+Bx+4\frac{1}{(x-4)(x+4)} = \frac{A}{x-4} + \frac{B}{x+4}.
  3. Solve for AA and BB: 1=A(x+4)+B(x−4)1 = A(x+4) + B(x-4).
  4. Let x=4  ⟹  A=18x=4 \implies A = \frac{1}{8}. Let x=−4  ⟹  B=−18x=-4 \implies B = -\frac{1}{8}.
  5. Substitute back: ∫(1/8x−4−1/8x+4)dx=18ln⁡∣x−4∣−18ln⁡∣x+4∣+C\int (\frac{1/8}{x-4} - \frac{1/8}{x+4}) dx = \frac{1}{8} \ln|x-4| - \frac{1}{8} \ln|x+4| + C.
  6. Simplify: 18ln⁡∣x−4x+4∣+C\frac{1}{8} \ln \left| \frac{x-4}{x+4} \right| + C.

Explanation:

This example demonstrates how a quadratic denominator is split into two linear factors to facilitate logarithmic integration.

Problem 4:

Evaluate ∫xexdx\int x e^x dx using integration by parts.

Graph showing the product function x times e^x which requires integration by parts.

Solution:

  1. Choose u=xu = x (algebraic) and dv=exdxdv = e^x dx (exponential) based on ILATE.
  2. Compute du=dxdu = dx and v=∫exdx=exv = \int e^x dx = e^x.
  3. Apply the formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du.
  4. ∫xexdx=xex−∫exdx\int x e^x dx = x e^x - \int e^x dx.
  5. Evaluate the remaining integral: xex−ex+Cx e^x - e^x + C.
  6. Factor out exe^x: ex(x−1)+Ce^x(x-1) + C.

Explanation:

Integration by parts is used here to reduce the algebraic term xx to 11 through differentiation, leaving a simple exponential integral.