Question

Find `int 1/(x^2+2x+10)^2 dx`?

Answer 1

`1/54{arc tan((x+1)/3)+(3(x+1))/(x^2+2x+10)}+C`.

Explanation:

Let, `I=int1/(x^2+2x+10)^2dx=int1/{(x+1)^2+9}^2dx`.

We use the subst., `(x+1)=3tanu rArr dx=3sec^2udu`.

`:. I=int1/(9tan^2u+9)^2*3sec^2udu`,

`=int(3sec^2u)/(81sec^4u)du`,

`=1/27int1/sec^2udu=1/27intcos^2udu`,

`=1/27int(1+cos2u)/2du`,

`=1/54(u+1/2*sin2u)`,

`=1/54{u+1/2*(2tanu)/(1+tan^2u)}`,

`=1/54{u+tanu/(1+tan^2u)}`.

Replacing `u" by "arc tan((x+1)/3)`, we finally have,

`I=1/54[arc tan((x+1)/3)+{(x+1)/3}/{1+(x+1)^2/3^2}]`.

`rArr I=1/54{arc tan((x+1)/3)+(3(x+1))/(x^2+2x+10)}+C`.

Enjoy Maths., and, Spread The Joy!

Answer 2

`int \ 1/(x^2+2x+10)^2 \ dx = 1/54 arctan((x+1)/3)+(x+1)/(18(9+(x+1)^2)) + C`

Explanation:

We seek:

`I = int \ 1/(x^2+2x+10)^2 \ dx`

We complete the square on the quadratic in the denominator:

`I = int \ 1/((x+1)^2-1+10)^2 \ dx`
`\ \ = int \ 1/((x+1)^2+9)^2 \ dx`

Now we can attempt a substitution of the form:

`x+1 = 3tan theta => tan theta = (x+1)/3`
And, `dx/(d theta) = 3sec^2 theta`

If we substitute into the integral, it becomes:

`I = int \ 1/(9tan^2 theta+9)^2 \ 3sec^2 theta \ d theta`
`\ \ = int \ 1/((9(tan^2 theta + 1))^2) \ 3sec^2 theta \ d theta`
`\ \ = 1/27 \ int \ 1/((sec^2 theta)^2) \ sec^2 theta \ d theta`
`\ \ = 1/27 \ int \ 1/sec^2 theta \ d theta`
`\ \ = 1/27 \ int \ cos^2 theta \ d theta`

And if we apply the cosine double angle formula:

`cos 2A -= 2cos^2A-1`

We get:

`I = 1/27 \ int \ (1+cos 2theta)/2 \ d theta`
`\ \ = 1/54 \ int \ 1+cos 2theta \ d theta`
`\ \ = 1/54 \ {theta+1/2sin2theta } + C`

We now manipulate the result in preparation of the restoration of the earlier substitution:

`I = 1/54 \ {theta+1/2(2sin theta cos theta) } + C`
`\ \ = 1/54 \ {theta+sin theta cos theta * (cos theta)/(cos theta) } + C`
`\ \ = 1/54 \ {theta+(tan theta)/(sec^2 theta) } + C`
`\ \ = 1/54 \ {theta+(tan theta)/(1+tan^2 theta) } + C`

And now, we can restore the earlier substitution:

`I = 1/54 \ {arctan((x+1)/3)+((x+1)/3)/(1+((x+1)/3)^2) } + C`
`\ \ = 1/54 \ {arctan((x+1)/3)+((x+1)/3)/(1+(x+1)^2/9) } + C`
`\ \ = 1/54 \ {arctan((x+1)/3)+((x+1)/3)/((9+(x+1)^2)/9) } + C`
`\ \ = 1/54 \ {arctan((x+1)/3)+(3(x+1))/(9+(x+1)^2) } + C`
`\ \ = 1/54 arctan((x+1)/3)+(x+1)/(18(9+(x+1)^2)) + C`