Integral help

My solution

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2. \int x^2sinxdx ?

Formula:

$\begin{array}{|rcll|} \hline \int(u\cdot v')\ dx = u\cdot v - \int(u'\cdot v)\ dx \\ \hline \end{array}$

$\begin{array}{|rclrclrcl|} \hline && \int (~x^2\cdot \sin(x) ~)\ dx\\ && & u &=& x^2 & u' &=& 2x \\ && & v' &=& \sin(x) & v &=& -\cos(x) \\ &=& -x^2\cdot \cos(x) + 2\cdot\int (x\cdot \cos(x) )\ dx \\ \\ \hline && \int (~x\cdot \cos(x) ~)\ dx\\ && & u &=& x & u' &=& 1 \\ & & & v' &=& \cos(x) & v &=& \sin(x) \\ &=& x\cdot \sin(x) - \int (1\cdot \sin(x) )\ dx \\ &=& x\cdot \sin(x) - \int (\sin(x) )\ dx & \int \sin(x) &=& -\cos(x) \\ &=& x\cdot \sin(x) + \cos(x) \\ \\ \hline && \mathbf{ \int (~x^2\cdot \sin(x) ~)\ dx }\\ &=& -x^2\cdot \cos(x) + 2\cdot[x\cdot \sin(x) + \cos(x) ] + c \\ &=& -x^2\cdot \cos(x) + 2\cdot x\cdot \sin(x)+ 2\cos(x)+ c \\ &=& \mathbf{ 2\cdot x\cdot \sin(x) -(x^2-2)\cdot \cos(x) + c } \\ \hline \end{array}$

1)

Take the integral:
 integral x/(x^2 + 6 x + 10) dx
Rewrite the integrand x/(x^2 + 6 x + 10) as (2 x + 6)/(2 (x^2 + 6 x + 10)) - 3/(x^2 + 6 x + 10):
 = integral((2 x + 6)/(2 (x^2 + 6 x + 10)) - 3/(x^2 + 6 x + 10)) dx
Integrate the sum term by term and factor out constants:
 = 1/2 integral(2 x + 6)/(x^2 + 6 x + 10) dx - 3 integral1/(x^2 + 6 x + 10) dx
For the integrand (2 x + 6)/(x^2 + 6 x + 10), substitute u = x^2 + 6 x + 10 and du = (2 x + 6) dx:
 = 1/2 integral1/u du - 3 integral1/(x^2 + 6 x + 10) dx
The integral of 1/u is log(u):
 = (log(u))/2 - 3 integral1/(x^2 + 6 x + 10) dx
For the integrand 1/(x^2 + 6 x + 10), complete the square:
 = (log(u))/2 - 3 integral1/((x + 3)^2 + 1) dx
For the integrand 1/((x + 3)^2 + 1), substitute s = x + 3 and ds = dx:
 = (log(u))/2 - 3 integral1/(s^2 + 1) ds
The integral of 1/(s^2 + 1) is tan^(-1)(s):
 = (log(u))/2 - 3 tan^(-1)(s) + constant
Substitute back for s = x + 3:
 = (log(u))/2 - 3 tan^(-1)(x + 3) + constant
Substitute back for u = x^2 + 6 x + 10:
Answer: |= 1/2 log(x^2 + 6 x + 10) - 3 tan^(-1)(x + 3) + constant

2)

Take the integral:
 integral x^2 sin(x) dx
For the integrand x^2 sin(x), integrate by parts, integral f dg = f g - integral g df, where
 f = x^2, dg = sin(x) dx, df = 2 x dx, g = -cos(x):
 = -x^2 cos(x) + 2 integral x cos(x) dx
For the integrand x cos(x), integrate by parts, integral f dg = f g - integral g df, where
 f = x, dg = cos(x) dx, df = dx, g = sin(x):
 = -x^2 cos(x) + 2 x sin(x) - 2 integral sin(x) dx
The integral of sin(x) is -cos(x):
 = x^2 (-cos(x)) + 2 x sin(x) + 2 cos(x) + constant
Which is equal to:
Answer: |= 2 x sin(x) - (x^2 - 2) cos(x) + constant

int of x^2sinx dx

$\int x^2sinx\;dx\\ \qquad \text{Integrate by parts}\\ \qquad u=x^2\qquad v'=sinx\\ \qquad u'=2x\qquad v=-cosx\\ =[-x^2cosx]-\int -2xcosx\;dx\\ =[-x^2cosx]+2\int xcosx\;dx\\ \qquad \text{Integrate by parts}\\ \qquad u=x\qquad v'=cosx\\ \qquad u'=1\qquad v=sinx\\ =[-x^2cosx]+2\{(xsinx)-\int sinx\;dx\}\\ =[-x^2cosx]+2\{xsinx+cosx\;\}+c\\ =-x^2cosx+2xsinx+2cosx+c\\ =(2-x^2)cosx+2xsinx+c$

1)

$\int\;\frac{x}{x^2+6x+10}\;dx\\ =\frac{1}{2}\int\;\frac{2x+6}{x^2+6x+10}\;dx-\int\;\frac{-3}{x^2+6x+10}\;dx\\ =\frac{ln|x^2+6x+10|}{2}+3\int\;\frac{1}{x^2+6x+9+1}\;dx\\ =\frac{ln|x^2+6x+10|}{2}+3\int\;\frac{1}{(x+3)^2+1}\;dx\\ =\frac{ln|x^2+6x+10|}{2}+3tan^{-1}(x+3)+c\\$