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Integrals Maths Integrals

Kindly Consider the following

107. $\frac{x + 3}{x^{2} - 2 x - 5}$

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Vishal Baghel

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$\begin{array}{l} L e t (x + 3) = A \frac{d}{d x} (x^{2} - 2 x - 5) + B \\ \Rightarrow (x + 3) = A (2 x - 2) + B \\ E q u a t i n g t h e c o e f f i c i e n t o f x a n d c o n s t a n t t e r m o n b o t h s i d e, w e g e t \\ W e g e t 2 A = 1 a n d - 2 A + B = 3 \\ A = \frac{1}{2} B = 4 \\ T h e r e f o r e, (x + 3) = \frac{1}{2} (2 x - 2) + 4 - - - - - (1) \\ I = \int \frac{x + 3}{x^{2} - 2 x - 5} d x = \int \frac{\frac{1}{2} (2 x - 2) + 4}{x^{2} - 2 x - 5} d x \\ = \frac{1}{2} \int \frac{2 x - 2}{x^{2} - 2 x - 5} d x + 4 \int \frac{1}{x^{2} - 2 x + 5} d x \\ I = \int \frac{x + 3}{x^{2} - 2 x - 5} d x = \frac{1}{2} I_{1} + 4 I_{2} \\ I_{1} = \int \frac{2 x - 2}{x^{2} - 2 x - 5} d x \\ \begin{array}{l} L e t x^{2} - 2 x - 5 = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \Rightarrow (2 x - 2) d x = d t \end{array} \\ I_{1} = \int \frac{d t}{t} = l o g | t | = l o g | x^{2} - 2 x - 5 | - - - - - (2) \\ I_{2} = \int \frac{1}{x^{2} - 2 x - 5} d x = \int \frac{1}{(x^{2} - 2 x + 1) - 6} d x = \int \frac{1}{{(x - 1)}^{2} - (\sqrt6)^{2}} d x \\ = \frac{1}{2√6} l o g (\frac{x - 1 -\sqrt6}{x - 1 +\sqrt6}) - - - - - (3) \\ S u b s t i t u t i n g (2) a n d (3) i n (1), w e g e t \\ I = \frac{1}{2} l o g | x^{2} - 2 x - 5 | + \frac{4}{2} l o g | \frac{x - 1 -\sqrt6}{x - 1 +\sqrt6} | + C \end{array}$ $\begin{array}{l} \end{array}$

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Integrals Maths Integrals

106. Kindly Consider the following

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Vishal Baghel

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Kindly go through the solution

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New answer posted

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Integrals Maths Integrals

105. Kindly Consider the following

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Vishal Baghel

Contributor-Level 10

Kindly go through the solution

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New answer posted

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Integrals Maths Integrals

104. Kindly Consider the following

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} \frac{6 x + 7}{x^{2} - 9 x + 20} \\ L e t 6 x + 7 = A \frac{d}{d x} (x^{2} - 9 x + 20) + B \\ \Rightarrow 6 x + 7 = A (2 x - 9) + B \\ E q u a t i n g t h e c o e f f i c i e n t o f x a n d c o n s t a n t t e r m, w e h a v e \\ \begin{array}{l} 2 A = 6 a n d - 9 A + B = 7 \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \Rightarrow A = 3 \Rightarrow B = 34 \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} ∴ 6 x + 7 = 3 (2 x - 9) + 34 \end{array} \end{array}$

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Integrals Maths Integrals

Kindly Consider the following

103. $\frac{5 x - 2}{1 + 2 x + 3 x^{2}}$

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} L e t 5 x - 2 = A \frac{d}{d x} (1 + 2 x + 3 x^{2}) + B \\ \Rightarrow 5 x - 2 = A (2 + 6 x) + B \\ = 2 A + A 6 x + B \end{array}$

$\begin{array}{l} E q u a t i n g t h e c o e f f i c i e n t o f x a n d c o n s t a n t t e r m o n b o t h s i d e s, w e h a v e \\ 5 = 6 A; a n d 2 A + B = - 2 \\ \Rightarrow A = \frac{5}{6} \\ \Rightarrow B = - 2 - 2 A \\ B = \frac{- 11}{3} \\ T h e r e f o r e, \\ 5 x - 2 = \frac{5}{6} (2 + 6 x) + (\frac{- 11}{3}) \\ ∴ I = \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x \\ = \int \frac{\frac{5}{6} (2 + 6 x) + (\frac{- 11}{3})}{1 + 2 x + 3 x^{2}} d x \\ = \frac{5}{6} \int \frac{(2 + 6 x)}{1 + 2 x + 3 x^{2}} d x - \frac{11}{3} \int \frac{1}{1 + 2 x + 3 x^{2}} d x \\ L e t, I_{1} = \int \frac{2 + 6 x}{1 + 2 x + 3 x^{2}} d x a n d I_{2} = \int \frac{1}{1 + 2 x + 3 x^{2}} d x \\ ∴ I = \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x \\ = \frac{5}{6} I_{1} - \frac{11}{3} I_{2} - - - - - (1) \\ \begin{array}{l} L e t 1 + 2 x + 3 x^{2} = t \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \Rightarrow (2 + 6 x) d x = d t \end{array} \\ I_{1} = \int \frac{d t}{t} = l o g | t | \\ = l o g | 1 + 2 x + 3 x^{2} |____(2) \\ I_{2} = \int \frac{1}{1 + 2 x + 3 x^{2}} d x \\ N o w, 1 + 2 x + 3 x^{2} = 1 + 3 (x^{2} + \frac{2}{3} x) \\ T h e r e f o r e, 1 + 3 (x^{2} + \frac{2}{3} x) = 1 + 3 (x^{2} + \frac{2}{3} x + \frac{1}{9} - \frac{1}{9}) \end{array}$