Question

evaluate indefinite integral using the substitution method ..........again lol

Answer 1

\[\int\limits_{}^{}x/\left( 1-x ^{^{2}} \right)^{-1/5}xdx\]

Answer 2

sorry not -1/5 its 1/5

Answer 3

i got \[-5/8\left( 1-x ^{2} \right)^{^{4/5}}+K\]

Answer 4

is this right?

Answer 5

Let \( 1 - x^2 = u\),then \( du = 2xdx\)

\( =\large \frac{1}{2} \int \frac{2x}{(1 - x^2)^{-1/5}} dx\)
\( =\large \frac{1}{2} \int \frac{du}{(u)^{-1/5}} dx\)
\( =\large\frac{1}{2} [\frac{u^{5/6}}{5/6}] + C \)
\( =\large\frac{3}{5} u^{5/6} + C \)

Answer 6

\( 3/5 (1-x^2)^{5/6}\)

Answer 7

wouldn't du=-2xdx?

Answer 8

im sorry and its not ^(-1/5) its positive

Answer 9

\[
\int \frac{x}{\sqrt[5]{1-x^2}}
\, dx=-\frac{5}{8}
\left(1-x^2\right)^{4/5}
\]
by following the method of @experimentX with the -1/5 replaced with 1/5

Answer 10

thanks

Answer 11

Sorry, there was mistake from second step

\( = \large -\frac{1}{2} [\frac{u^{4/5}}{4/5}] + C\)
\( = \large -\frac{5}{8} u^{4/5} + C\)
\( = \large -\frac{5}{8} (1-x^2)^{4/5} + C\)

Answer 12

omg thanks im so getting this now