Question

please integrate this #1.) x^3 squareroot of (a^2+x^2) dx
#2.)dx/x(x^2-a)^2
#3.)z^5 dz/(z^2-9^2)

Answer 1

For number 1, let u=x^3 so we can get rid of the x^3 to use the arcsin formula. The arcsin formula is integral of du/sqrt of a^2-u^2= arcsin u/a + c.

Answer 2

can i ask is the a here has a value?

Answer 3

the first one you should sub u=a^2+x^2

Answer 4

the other guy is wrong :|

Answer 5

then? can u plaease help me out?

Answer 6

\[\int\limits x^3 \sqrt{a^2 +x^2} dx \]

if its this

Answer 7

u=a^2 +x^2

du/dx= 2x

so x dx = du /2

Answer 8

rearranging the substitution , x^2 = u-a^2

Answer 9

rewrite the integral \[\int\limits x \times x^2 \times \sqrt{a^2+x^2} dx = \int\limits x^2 \times \sqrt{a^2+x^2} \times (x dx ) \]

Answer 10

is this integral by parts?

Answer 11

\[= \int\limits (u-a^2) \times \sqrt{u} \times \frac{du}{2} = \frac{1}{2} \int\limits (u^{\frac{3}{2}} - a^2 u^{\frac{1}{2}} )du \]

Answer 12

no by substitution

Answer 13

its easy from then ^

Answer 14

a^2 is just a constant

Answer 15

thank u. wat about the number 2 and 3?

Answer 16

3 is by partial fractions

Answer 17

wait, thats wrong

Answer 18

you need to divide first , because the degree of the numerator is larger than the denominator

Answer 19

its z^5 /(z^2-a^2)dz

Answer 20

the second one is partial fractions as well \[= \int\limits \frac{dx}{ x (x-\sqrt{a})^2 (x+\sqrt{a})^2} = \int\limits \frac{A}{x} + \frac{B}{x-\sqrt{a}} +\frac{C}{ (x-\sqrt{a})^2} + \frac{D}{(x+\sqrt{a})} + \frac{E}{(x+\sqrt{a})^2}\]