Find the indefinite and definite integrals (4 questions)?

Question

1) (indefinite) arctanx/(x^2+1) dx

2) (indefinite) [cubed root of (x^3 +1)]*x^5 dx

3)(indefinite) (4+x)/(x^2+1) dx

4) integral from 0 to 4: x*(sqrt(16-x^2)) dx

Ive tried all of these problems several times, for the past 3 hours & I still dont have the right answer --- help! Thanks!

Answer 1

2) Int [cubed root of (x^3 +1)]*x^5 dx
u = x^3 + 1
du = 3x^2 dx
du/3 = x^2 dx

u-1 = x^3
x^5 dx = (u - 1) du/3

Int [cubed root of (x^3 +1)]*x^5 dx
= Int [cubed root of u]*(u - 1) du/3
= Int [ u^(1/3) (u- 1) du/3
= Int [ u^(4/3) - u^(1/3) ] du/3
= (1/3)[ (3/7) u^(7/3) - (3/4)u^(4/3)] + C
= (1/3)[ (3/7) (x^3+1)^(7/3) - (3/4)(x^3+1)^(4/3)] + C

3) Int (4+x)/(x^2+1) dx
= Int { (4/(x^2+1) dx + Int {x/(x^2 + 1) dx

Int { (4/(x^2+1) dx = 4 Int { dx/(x^2+1^2)
Int { (4/(x^2+1) dx = 4 [ arctan x ] + C

u = x^2+1 du = 2x dx
du/2 = xdx

Int {x/(x^2 + 1) dx = Int { 1/u du/2
Int {x/(x^2 + 1) dx = 1/2 Int { 1/u du
Int {x/(x^2 + 1) dx = 1/2 [ ln u ] + C
Int {x/(x^2 + 1) dx = 1/2 [ ln (x^2+1) ] + C

Int { (4/(x^2+1) dx + Int {x/(x^2 + 1) dx
=4 [ arctan x ] + 1/2 [ ln (x^2+1) ] + C

4)
u = 16 - x^2
du = -2xdx
-du/2 = x dx

Int {x*(sqrt(16-x^2)) dx
= Int {(sqrt(16-x^2)) xdx
= Int { u^(1/2) * -du/2
= -½ [ (2/3)u^(3/2) ]

x=0 u=16
x=4 u= 0

-½ [ (2/3)u^(3/2) ] from 16 to 0
=-½ [ (2/3)(16)^(3/2) ]
= -64/3

Answer 2

For #4, make the substitution y = sqrt (16-x^2). The integral won't be much harder than integrating ydy.

For #3, split it into two fractions with numerators 4 and x. To integrate the latter, let y = the denominator. And I think you can do the former.

#1 is really easy too. Just let y = arctan x. You're integrating exactly y*dy

For #2 I'd let y = what's inside the radical. Then you have y^(1/3)*(y-1)dy, times a factor of 1/3. Split that into the obvious two terms, and it's easy too. :)

Answer 3

? x/(root(one million+2x^2)) dx positioned one million+2x^2 = t So 4x dx = dt ? x/(root(one million+2x^2)) dx =?t^(-one million/2) dt /4 = t^(one million/2) /2 Substituting lower back for t, ? x/(root(one million+2x^2)) dx = root(one million+2x^2) /2 ? (2-x)rootx dx = ? (2x^(one million/2) -x^(3/2) dx =(4/3)x^(3/2) - (2/5) x^(5/2) ? (x+one million)/(x^2+2x-3) dx Use the tactic of partial fractions right here. (x+one million)/(x^2+2x-3) = (x+one million)/{(x-3)(x+one million)} positioned (x+one million)/{(x-3)(x+one million)} = A/(x-3) + B/(x+one million) fixing we get A= -one million/2 and B= one million/2 So, ? (x+one million)/(x^2+2x-3) dx = ? (-one million/2)/(x-3) dx + ? (one million/2)/(x+one million) dx = (-one million/2) ln(x-3) +(one million/2) ln(x+one million)