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Published byMildred Bates Modified over 6 years ago

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Indefinite integrals Definition: if f(x) be any differentiable function of such that d/dx f( x ) = f(x)Is called an anti-derivative or an indefinite integral of f(x) we write f(x) = ∫ f (x) dx. The symbol ∫ dx is used as a symbol of operation which means integrating w.r.t. x.

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**1.∫ k f(x) dx = k ∫ f(x) dx, where k is a constant **

In all the above formulas, c is the constant of integration. Some standard results on integration 1.∫ k f(x) dx = k ∫ f(x) dx, where k is a constant 2. ∫ ( f(x) ± g(x))dx = ∫ f(x)dx ± ∫ g(x)dx

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Example: ∫4x⁷ dx =4∫x⁷ dx =4∫xⁿ⁺¹/n+1 =4 x⁸/ + c = x⁸/8 + c 8

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(1) ∫(sinx + cosx) dx = ∫sinx dx + ∫cosx dx = -cosx + sinx + C

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**METHODS OF INTEGRATION**

It was based on inspection, i.e,. On the search of a function F whose derivative is f which led us to the integral of f . However, this method, which depends on inspection , is not very suitable for many functions. Hence, we need to develop additional techniques or methods for finding the integrals by reducing Them into standard forms.

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**Following are the method**

of integration: Integration by substitution Integration by part Integration by partial fractions

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**Integration by substitution**

This method is used when basic rules of Integration given above, are not directly applicable. In such cases, the integration is transformed (by suitable substitution of given variable by a new variable) into a form to which basic rules of integration are applicable

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**∫√t dt = ∫t½ dt = 2/3t³⁄² + C Equation of integrals √15 + logx dx x**

put 15 + logx = t 1/x dx = dt Substituting in given question, we get ∫√t dt = ∫t½ dt = 2/3t³⁄² + C

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Conti…… = 2/3 (15 + logx)³⁄² + C Where c = Constant of integration.

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**∫ Integration by parts ∫ ∫ ∫ This method is based on product rule**

differentiation. According to this method, the integral of product of Two functions f(x). g(x) is given by ∫ g(x) dx - f ’(x) ( g(x)dx) dx ∫ ∫ ∫ F(x) . g(x) dx = f(x)

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**∫ I . II dx = I ∫IIdx - ∫(d(I)/dx) dx**

Taking f(x) as the first function (I) and g(x) as the second (II) the above equation can be Stated as ∫ I . II dx = I ∫IIdx - ∫(d(I)/dx) dx

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**Rule for the proper chose of the first function**

The first function to be selected may be the one which comes first in the order ILATE where I means inverse trigonometric function L means logarithmic function A means algebraic T means trigonometric E means exponential.

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∫x² e dx x x Take x² as the function and e as the second function and integrating by parts, we get I = x² e - ∫2x e x X dx

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**I = x² e – 2 [xe - ∫e dx] = x² e – 2xe + 2e + C,**

Again apply by parts in second term, we get x x I = x² e – 2 [xe - ∫e dx] X X = x² e – 2xe + 2e + C, Where c. is the constant of integration X

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**Integration by partial fraction**

If f(x) and g(x) are two polynomials, then f(x)/g(x) is called a rational algebraic Function of x. Any rational f (x)/ g(x) can be expressed as the sum of rational functions each having a simple factor of g(x). Each such fraction is called a partial fraction.

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3x 1. ∫ dx (x-1) (x+2) A B = + X-1 X + 2 3x = A (X + 2) + B (X + 1)

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Put,, X = we get 3 = 3A A = 1 Put, x = -2 we get - 6 = - 3B B = 2

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**∫ ∫ ∫ = (x+2) (x-1) 2 + 1 X - 1 3x X-1 X + 2 3x dx + 2 X + 2**

Now integrating, we get 3x 2 1 X - 1 ∫ ∫ dx = ∫ dx + 2 ( x - 1 ) ( x + 2) X + 2

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= log ( x – 1 ) + 2 log ( x + 2 ) + c

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Thank you Presented by, NUZHAT SHAHEEN

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