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  • Integrate the functions in Exercises 1 to 24. 6

Integrate the functions in Exercises 1 to 24.

    Q6.    \frac{5x}{(x+1)(x^2 + 9)}

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Let us assume that   :

                                               \frac{5x}{(x+1)(x^2 + 9)}\ =\ \frac{A}{(x+1)}\ +\ \frac{Bx + c}{x^2 + 9}

Solving the equation and comparing coefficients of x2, x and the constant term.

We get,

                                                A\ =\ \frac{-1}{2}\ ;\ B\ =\ \frac{1}{2}\ ;\ C\ =\ \frac{9}{2}

Thus the equation becomes :

                                                 \frac{5x}{(x+1)(x^2 + 9)}\ =\ \frac{-1}{2(x+1)}\ +\ \frac{\frac{x}{2}+\frac{9}{2}}{x^2 + 9}

or                                           

                                                 \int \frac{5x}{(x+1)(x^2 + 9)}\ =\ \int \left [ \frac{-1}{2(x+1)}\ +\ \frac{x+9}{2(x^2 + 9}) \right ]dx

or                                                                                         =\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{2} \int \frac{x}{x^2 +9}dx +\frac{9}{2} \int \frac{1}{x^2+9}dx

or                                                                                         =\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{4} \int \frac{2x}{x^2 +9}dx +\frac{9}{2} \int \frac{1}{x^2+9}dx

or                                                                                         =\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{4} \log {(x^2 +9)} +\frac{3}{2} \tan^{-1}\frac{x}{3}\ +\ C

Posted by

Devendra Khairwa

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