How to integrate &#x222B; 1 x 2 ( 2 x - 1 ) using partial fractions?

Question

How to integrate           &amp;#x222B;              1                              x            2                                (            2            x            -            1            )                               using partial fractions?

Peruvianoe4p · Accepted Answer

We need to find         A     ,     B     ,     C     such that                1                           x           2                             (           2           x           -           1           )                          =            A       x          +            B                x         2                 +            C                2         x         -         1                 for all         x    . Multiply both sides by                x       2                 (       2       x       -       1       )          to get         1     =     A     x            (       2       x       -       1       )          +     B            (       2       x       -       1       )          +     C            x       2                  1     =     2     A            x       2          -     A     x     +     2     B     x     -     B     +     C            x       2                  1     =            (       2       A       +       C       )                 x       2          +            (       2       B       -       A       )          x     -     B     Equating coefficients give us                {                                        2             A             +             C             =             0                                                     2             B             -             A             =             0                                                     -             B             =             1                                     And thus we have         A     =     -     2     ,     B     =     -     1     ,     C     =     4    . When this is substituted in the initial equation, we obtain                1                           x           2                             (           2           x           -           1           )                          =            4                2         x         -         1                 -            2       x          -            1                x         2                 Now, integrate it term by term         &amp;#x222B;     &amp;#xA0;            4                2         x         -         1                 &amp;#xA0;            d       x          -     &amp;#x222B;     &amp;#xA0;            2       x          &amp;#xA0;            d       x          -     &amp;#x222B;     &amp;#xA0;            1                x         2                 &amp;#xA0;            d       x          to get         2     ln            |       2       x       -       1       |          -     2     ln            |       x       |          +            1       x          +     C

davz198888za · Accepted Answer

Perform the decomposition into partial fractions

\frac{1}{x^{2} (2 x - 1)} = \frac{A}{x^{2}} + \frac{B}{x} + \frac{C}{2 x - 1}

= \frac{A (2 x - 1) + B x (2 x - 1) + C (x^{2})}{x^{2} (2 x - 1)}

Compare the numerators because the denominators are the same.

1 = A (2 x - 1) + B x (2 x - 1) + C (x^{2})

Let

x = 0

,

\Rightarrow

,

1 = - A

,

\Rightarrow

,

A = - 1

Let

x = \frac{1}{2}

,

\Rightarrow

,

1 = \frac{C}{4}

,

\Rightarrow

,

C = 4

Coefficients of

x^{2}

0 = 2 B + C

B = - \frac{C}{2} = - \frac{4}{2} = - 2

Hence,

\frac{1}{x^{2} (2 x - 1)} = - \frac{1}{x^{2}} - \frac{2}{x} + \frac{4}{2 x - 1}

So,

\int \frac{1 d x}{x^{2} (2 x - 1)} = - \int \frac{1 d x}{x^{2}} - \int \frac{2 d x}{x} + \int \frac{4 d x}{2 x - 1}

= \frac{1}{x} - 2 \ln (| x |) + 2 \ln (| 2 x - 1 |) + C

How to integrate int 1/(x^2(2x-1)) using partial fractions?

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