Question

Need serious help with integration by parts. Typing Question now.......

Answer 1

\[\int\limits_{}^{} 64x ^{2}*\ln(x) dx\]

Answer 2

I just cant seem to do integration by parts, if some can please help, that would be great.

Answer 3

ln(x) should be your downward

Answer 4

\[\int{64x^2}{ln(x)}dx=\frac{64x^3}{3}ln(x)-\int\frac{64x^3}{3x}dx\]

Answer 5

\[\int{64x^2}{ln(x)}dx=\frac{64x^3}{3}ln(x)-\int\frac{64x^3}{3x}dx\]
\[=\frac{64x^3}{3}ln(x)-\int\frac{64x^2}{3}dx\]
\[=\frac{64x^3}{3}ln(x)-\frac{64x^3}{9}+C\]
maybe?

Answer 6

\[=\frac{64x^3}{3}ln(x)-\frac{64x^3}{9}+C\]
\[=\frac{64x^3}{9}(3ln(x)-1)+C\] is what the wolf shows

Answer 7

i get the first term, but for the second i get (64/6)*x^2

Answer 8

lol wolf is the man

Answer 9

3*3 = 9, not 6

Answer 10

\[\int udv=uv-\int vdu\]
\[\begin{array}c
u=ln(x)&du=\frac{1}{x}dx\\
dv=64x^2& v= \frac{64}{3}x^3
\end{array}\]
\[\int ln(x)\ [64x^2\ dx]=ln(x)\ \frac{64}{3}x^3 -\int\frac{64}{3}x^3[\frac{1}{x} dx] \]
\[\int ln(x)\ 64x^2\ dx=ln(x)\ \frac{64}{3}x^3 -\int\frac{64}{3}x^2 dx \]
\[\int ln(x)\ 64x^2\ dx=ln(x)\ \frac{64}{3}x^3 -\frac{64}{3}\frac{1}{3}x^3 +C \]

Answer 11

Hey i get it now. Calc makes my head hurt.

Answer 12

recall product rule:
\[(fg)'=f'g+fg'\]
integrating both sides we get
\[\int\limits_{}^{}(fg)' dx=\int\limits_{}^{}[f'g +fg'] dx=\int\limits_{}^{} f' g dx+\int\limits_{}^{} fg' dx\]
so we have
\[\int\limits_{}^{}(fg)' dx=\int\limits_{}^{}f' g dx+\int\limits_{}^{}f g' dx\]
but \[\int\limits_{}^{}(fg)' dx=fg+C \] but for now we will just avoid the C and write
\[\int\limits_{}^{}(fg)' dx=fg\]
so we have
\[fg=\int\limits_{}^{} f' g dx+\int\limits_{}^{} f g' dx\]
now if we solve for \[\int\limits_{}^{}f'g dx\] we get the formula for integration by parts (you could have chose to solve the other one instead)
\[\int\limits_{}^{}f' g dx=fg - \int\limits_{}^{}fg' dx\]

Answer 13

make a 7 and you have you integration by parts
i remebered it by " across i multiply(x) and then i go across diagonal and down=-X(-multiplication)