Question

How do i integrate:

Integral x^3sqrt(x^2+7)

Answer 1

\[\int\limits_{}^{}x^3\sqrt{x^2+7}\]

Answer 2

Change the square root symbol to power half and integrate like normal. Do you need help with integrating after this step?

x^2*(x^2+7)^(1/2)

Answer 3

the integral is x^3sqrt(x^2+7)

Answer 4

yes i still need help

Answer 5

Square root and to the power half is the same thing - try it on your calculator.

\[x ^{3}\sqrt{x^2+7} = x^3 (x^2+7)^{1/2}\]
\[\int\limits_{ }^{ }x^3 (x^2+7)^{1/2}=(1/15)(x ^{2}+7)^{3/2}(3x^2-14) +c\]

Answer 6

you just game the answer, not how to do it

Answer 7

what should i let u be to solve this?

Answer 8

(x^2 + 7)^(1/2)

Answer 9

that would serve no purpose as du/dx wuld then be: 1/2(x^2+7)^-1/2 *(2x)

Answer 10

Then I guess your only option is x^3. ;)

Answer 11

you dont know how to solve this do you

Answer 12

hey zip..i haven't worked this out completely...but i would thinnk let u = x^2 +7

Answer 13

yeah, that seems reasonable

Answer 14

I know perfectly well how to integrate by parts - but you seem like a master.

Answer 15

du would then be 1/2 du = xdx

Answer 16

come on dont pretend chaise

Answer 17

right,that would get rid of one of the x's, but then i think i might need to rewrite x in terms of u

Answer 18

i think you'll have to integrate a 2nd time b/c you'll have an x^2 left over

Answer 19

again...this is completely by observation...i've not worked anythign out

Answer 20

that makes sense

Answer 21

help please

Answer 22

\[\int x^3\sqrt{x^2+7}\,dx\quad,\quad y=x^2+7\quad,\quad dy=2xdx\]
\[\frac{1}{2}\int(y-7)\sqrt{y}\,dy=\frac{1}{2}\int(y^{3/2}-7y^{1/2})\,dy=\frac{1}{5}y^{5/2}-\frac{7}{3}y^{3/2}+C=\]
\[=y^{3/2}\left(\frac{1}{5}y-\frac{7}{3}\right)+C=\frac{1}{15}y^{3/2}\left(3y-35\right)+C=\]
\[\frac{1}{15}(x^2+7)^{3/2}\left(3(x^2+7)-35\right)+C=\frac{1}{15}(x^2+7)^{3/2}\left(3x^2-14\right)+C\]