Question

Calculate this integral \[\int\limits_{0}^{1} \frac{ x^2 +1 }{ x^3 +3x+2 } d \theta\]

Answer 1

partial fractions
\[ \frac{x^2+1}{ (x+1)(x+2)} = \frac{A}{(x+1)} + \frac{B}{x+2}\]

Answer 2

\[ A(x+2) + B(x+1) = x^2+1

Answer 3

\[ A(x+2) + B(x+1) = x^2+1\]

Answer 4

\[ x = -2\]
\[ A(-2+2)+ B(-2+1) = (-2)^2+1)\]
\[-B = 5 ->> B = -5\]

Answer 5

\[ x = -1 \]
\[ A(-1+2) + B(-1+1) = (-1)^2+1\]
\[ A = 2\]

Answer 6

therefore, \[ \frac{2}{(x+1)} - \frac{5}{(x+2)}\]

Answer 7

you can integate it now.

Answer 8

\[ 2\int \frac{1}{(x+2)} dx - 5\int\frac{1}{x+2)}dx\]

Answer 9

I'm not sure this works, I think maybe you couldn't see that the power that the x on the bottom row is raised to is 3 not 2, so can't factorise it into x+1 and x+2

Answer 10

oh pellet.

Answer 11

sorry!

Answer 12

long division

Answer 13

nevermind
u = x^3+ 3x +2

Answer 14

du / dx = 3x^2 + 3

Answer 15

\[ \int \frac{x^2+1}{x^2+3x+2}dx\]
\[ \int \frac{x^2+1}{u}* \frac{du}{3(x^2+1)}\]
\[ = \frac{1}{3} \int \frac{1}{u}du\]

Answer 16

\[ = \frac{1}{3} ln(u) +x\]

Answer 17

\[ \frac{1}{3} ln(3x^2+3) + C \]

Answer 18

the x, isnt ment to be there in the second one.

Answer 19

then sub in the limits.