Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.
(x+1)^3/(x^2)dx
(x+1)/(x^2) dx
1/(1+4(x^2)) dx
x(x-10)^10 dx
(cosx)/(1+(sinx)^2) dx
(1+tanx)^3/(cosx)^2
Thanks
Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.
(x+1)^3/(x^2)dx
(x+1)/(x^2) dx
1/(1+4(x^2)) dx
x(x-10)^10 dx
(cosx)/(1+(sinx)^2) dx
(1+tanx)^3/(cosx)^2
Thanks
It's not clear what you did here. If you want to integrate $\displaystyle \int\frac{(x+1)^3}{x^2}dx$, you can just expand the numerator $\displaystyle \int\frac{x^3+3x^2+3x+1}{x^2}dx$ and then split it up into a sum. Then you should be able to apply the general formula for polynomial integrands.
The first two have been taken care of.
for $\displaystyle \int{\frac{1}{1 + 4x^2}\,dx}$ you do not use a $\displaystyle u$ substitution. You use a trigonometric substitution.
$\displaystyle \int{\frac{1}{1 + 4x^2}\,dx} = \int{\frac{1}{4\left(\frac{1}{4} + x^2\right)}\,dx}$
$\displaystyle = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx}$.
Now use the substitution $\displaystyle x = \frac{1}{2}\tan{\theta}$.
Note that $\displaystyle \frac{dx}{d\theta} = \frac{1}{2}\sec^2{\theta}$.
So $\displaystyle dx = \frac{1}{2}\sec^2{\theta}\,d\theta$.
Therefore
$\displaystyle = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx} = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + \left(\frac{1}{2}\tan{\theta}\right)^2}\,\frac{1}{ 2}\sec^2{\theta}\,d\theta}$
$\displaystyle = \frac{1}{8}\int{\frac{1}{\frac{1}{4}(1 + \tan^2{\theta})}\,\sec^2{\theta}\,d\theta}$
$\displaystyle = \frac{1}{2}\int{\frac{1}{1 + \tan^2{\theta}}\,\sec^2{\theta}\,d\theta}$
$\displaystyle = \frac{1}{2}\int{\frac{1}{\sec^2{\theta}}\,\sec^2{\ theta}\,d\theta}$
$\displaystyle = \frac{1}{2}\int{1\,d\theta}$
$\displaystyle = \frac{1}{2}\theta + C$.
Now remembering that $\displaystyle x = \frac{1}{2}\tan{\theta}$
$\displaystyle 2x = \tan{\theta}$
$\displaystyle \theta = \arctan{2x}$.
So finally our answer is $\displaystyle \frac{1}{2}\arctan{2x} + C$.
For $\displaystyle \int{x(x - 10)^{10}\,dx}$ you could expand everything, but that could get messy.
Instead, integration by parts will need to be used.
Let $\displaystyle u = x$ so that $\displaystyle du = 1$.
Let $\displaystyle dv = (x - 10)^{10}$ so that $\displaystyle v = \int{(x - 10)^{10}\,dx} = \frac{1}{11}(x - 10)^{11}$.
Therefore $\displaystyle \int{x(x - 10)^{10}\,dx} = \frac{1}{11}x(x - 10)^{11} - \int{\frac{1}{11}(x - 10)^{11}\,dx}$
$\displaystyle = \frac{1}{11}x(x - 10)^{11} - \frac{1}{132}(x - 10)^{12} + C$
$\displaystyle = \frac{1}{11}(x - 10)^{11}\left[x - \frac{1}{12}(x - 10)\right] + C$
$\displaystyle = \frac{1}{11}(x - 10)^{11}\left(x - \frac{1}{12}x + \frac{5}{6}\right)$
$\displaystyle = \frac{1}{11}(x - 10)^{11}\left(\frac{11}{12}x + \frac{5}{6}\right)$
$\displaystyle = \frac{1}{132}(x - 10)^{11}(11x + 10)$.
For $\displaystyle \int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx}$, note that
$\displaystyle \frac{1}{\cos^2{x}} = \sec^2{x}$.
So $\displaystyle \int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx} = \int{(1 + \tan{x})\sec^2{x}\,dx}$
Now let $\displaystyle u = 1 + \tan{x}$ so that $\displaystyle \frac{du}{dx} = \sec^2{x}$.
The integral becomes
$\displaystyle \int{u\,\frac{du}{dx}\,dx}$
$\displaystyle = \int{u\,du}$
$\displaystyle = \frac{1}{2}u^2 + C$
$\displaystyle = \frac{1}{2}(1 + \tan{x})^2 + C$.