x-3-2x-1-dx-

Question Number 91542 by jagoll last updated on 01/May/20

\int \frac{x^{3}}{2 x + 1} d x = ?

Commented by Prithwish Sen 1 last updated on 01/May/20

put 2 x + 1 = \frac{1}{t} \Rightarrow dx = - \frac{dt}{{2 t}^{2}}

= - \frac{1}{16} \int \frac{{(1 - t)}^{3}}{t^{4}} dt

= \frac{1}{48} {(2 x + 1)}^{3} - \frac{3}{32} {(2 x + 1)}^{2} + \frac{3}{16} (2 x + 1) - \frac{1}{16} \ln (2 x + 1) + C

C = costant . please check

sorry sir . I fix it . Thank you .

Commented by jagoll last updated on 01/May/20

w r o n g i n \ln (2 x + 1)

it should be - \frac{1}{16} \ln (2 x + 1)

Commented by john santu last updated on 01/May/20

i n o t h e r m e t h o d b y l o n g d i v i s i o n

t h e n i n t e g r a t e t h e r e s u l t

Answered by mr W last updated on 01/May/20

\int \frac{x^{3}}{2 x + 1} d x

= \frac{1}{8} \int \frac{{(2 x)}^{3} + 1 - 1}{2 x + 1} d x

= \frac{1}{8} \int {\frac{(2 x + 1) (4 x^{2} - 2 x + 1)}{2 x + 1} - \frac{1}{2 x + 1}} d x

= \frac{1}{8} \int {4 x^{2} - 2 x + 1 - \frac{1}{2 x + 1}} d x

= \frac{1}{8} {\frac{4 x^{3}}{3} - x^{2} + x - \frac{1}{2} \ln (2 x + 1)} + C

Commented by Prithwish Sen 1 last updated on 01/May/20

Excellent

Commented by jagoll last updated on 01/May/20

y e s . v e r y c r e a t i v e s i r

Answered by jagoll last updated on 01/May/20

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