`int sqrt(x^2 + 2x) dx` Evaluate the integral

Sunday, October 18, 2009

$\displaystyle\int\sqrt{{{{x}}^{{2}}+{2}{x}}}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle\int\sqrt{{{{x}}^{{2}}+{2}{x}}}{\left.{d}{x}\right.}$

Let's rewrite the integrand by completing the square for $\displaystyle{{x}}^{{2}}+{2}{x}$ ,

$\displaystyle=\int\sqrt{{{{\left({x}+{1}\right)}}^{{2}}-{1}}}{\left.{d}{x}\right.}$

Now apply integral substitution,

Let u=x+1,

$\displaystyle\Rightarrow{d}{u}={1}{\left.{d}{x}\right.}$

$\displaystyle=\int\sqrt{{{{u}}^{{2}}-{1}}}{d}{u}$

Now again apply the integral substitution,

Let u=sec(v),

$\displaystyle\Rightarrow{d}{u}={\sec{{\left({v}\right)}}}{\tan{{\left({v}\right)}}}{d}{v}$

$\displaystyle=\int\sqrt{{{{\sec}}^{{2}}{\left({v}\right)}-{1}}}{\sec{{\left({v}\right)}}}{\tan{{\left({v}\right)}}}{d}{v}$

Now use the identity: $\displaystyle{{\sec}}^{{2}}{\left({x}\right)}={1}+{{\tan}}^{{2}}{\left({x}\right)}$

$\displaystyle=\int\sqrt{{{1}+{{\tan}}^{{2}}{\left({v}\right)}-{1}}}{\sec{{\left({v}\right)}}}{\tan{{\left({v}\right)}}}{d}{v}$

$\displaystyle=\int\sqrt{{{{\tan}}^{{2}}{\left({v}\right)}}}{\sec{{\left({v}\right)}}}{\tan{{\left({v}\right)}}}{d}{v}$

assuming $\displaystyle{\tan{{\left({v}\right)}}}\ge{0},\sqrt{{{{\tan}}^{{2}}{\left({v}\right)}}}={\tan{{\left({v}\right)}}}$

$\displaystyle=\int{{\tan}}^{{2}}{\left({v}\right)}{\sec{{\left({v}\right)}}}{d}{v}$

Using the identity: $\displaystyle{{\tan}}^{{2}}{\left({x}\right)}={{\sec}}^{{2}}{\left({x}\right)}-{1}$

$\displaystyle=\int{\left({{\sec}}^{{2}}{\left({v}\right)}-{1}\right)}{\sec{{\left({v}\right)}}}{d}{v}$

$\displaystyle=\int{\left({{\sec}}^{{3}}{\left({v}\right)}-{\sec{{\left({v}\right)}}}\right)}{d}{v}$

Apply the sum rule,

$\displaystyle=\int{{\sec}}^{{3}}{\left({v}\right)}{d}{v}-\int{\sec{{\left({v}\right)}}}{d}{v}$

Now let's evaluate the first integral by applying the integral reduction,

$\displaystyle\int{{\sec}}^{{n}}{\left({x}\right)}=\frac{{{{\sec}}^{{{n}-{1}}}{\left({x}\right)}{\sin{{\left({x}\right)}}}}}{{{n}-{1}}}+\frac{{{n}-{2}}}{{{n}-{1}}}\int{{\sec}}^{{{n}-{2}}}{\left({x}\right)}{\left.{d}{x}\right.}$

$\displaystyle\int{{\sec}}^{{3}}{\left({v}\right)}{d}{v}=\frac{{{{\sec}}^{{2}}{\left({v}\right)}{\sin{{\left({v}\right)}}}}}{{2}}+\frac{{{3}-{2}}}{{{3}-{1}}}\int{\sec{{\left({v}\right)}}}{d}{v}$

$\displaystyle\int{{\sec}}^{{3}}{\left({v}\right)}{d}{v}=\frac{{{{\sec}}^{{2}}{\left({v}\right)}{\sin{{\left({v}\right)}}}}}{{2}}+\frac{{1}}{{2}}\int{\sec{{\left({v}\right)}}}{d}{v}$

Now use the common integral: $\displaystyle\int{\sec{{\left({x}\right)}}}{\left.{d}{x}\right.}={\ln}{\left|{\sec{{\left({x}\right)}}}+{\tan{{\left({x}\right)}}}\right|}$

$\displaystyle\int{{\sec}}^{{3}}{\left({v}\right)}{d}{v}=\frac{{{{\sec}}^{{2}}{\left({v}\right)}{\sin{{\left({v}\right)}}}}}{{2}}+\frac{{1}}{{2}}{\left({\ln}{\left|{\sec{{\left({v}\right)}}}+{\tan{{\left({v}\right)}}}\right|}\right)}$

Now plug back the above integral and the common integral,

$\displaystyle=\frac{{{{\sec}}^{{2}}{\left({v}\right)}{\sin{{\left({v}\right)}}}}}{{2}}+\frac{{1}}{{2}}{\left({\ln}{\left|{\sec{{\left({v}\right)}}}+{\tan{{\left({v}\right)}}}\right|}\right)}-{\ln}{\left|{\sec{{\left({v}\right)}}}+{\tan{{\left({v}\right)}}}\right|}$

$\displaystyle=\frac{{{{\sec}}^{{2}}{\left({v}\right)}{\sin{{\left({v}\right)}}}}}{{2}}-\frac{{1}}{{2}}{\left({\ln}{\left|{\sec{{\left({v}\right)}}}+{\tan{{\left({v}\right)}}}\right|}\right)}$

$\displaystyle=\frac{{\sin{{\left({v}\right)}}}}{{{2}{{\cos}}^{{2}}{\left({v}\right)}}}-\frac{{1}}{{2}}{\left({\ln}{\left|{\sec{{\left({v}\right)}}}+{\tan{{\left({v}\right)}}}\right|}\right)}$

$\displaystyle=\frac{{{\sec{{\left({v}\right)}}}{\tan{{\left({v}\right)}}}}}{{2}}-\frac{{1}}{{2}}{\left({\ln}{\left|{\sec{{\left({v}\right)}}}+{\tan{{\left({v}\right)}}}\right|}\right)}$

Now substitute back: $\displaystyle{u}={\sec{{\left({v}\right)}}}.{u}={\left({x}+{1}\right)}$

$\displaystyle\Rightarrow{v}={a}{r}{c}{\sec{{\left({u}\right)}}}$

$\displaystyle\Rightarrow{v}={a}{r}{c}{\sec{{\left({x}+{1}\right)}}}$

$\displaystyle=\frac{{{\sec{{\left({a}{r}{c}{\sec{{\left({x}+{1}\right)}}}\right)}}}{\tan{{\left({a}{r}{c}{\sec{{\left({x}+{1}\right)}}}\right)}}}}}{{2}}-\frac{{1}}{{2}}{\left({\ln}{\left|{\sec{{\left({a}{r}{c}{\sec{{\left({x}+{1}\right)}}}\right)}}}+{\tan{{\left({a}{r}{c}{\sec{{\left({x}+{1}\right)}}}\right)}}}\right|}\right)}$

Now $\displaystyle{\tan{{\left({a}{r}{c}{\sec{{\left({x}+{1}\right)}}}\right)}}}=\sqrt{{{{\left({x}+{1}\right)}}^{{2}}-{1}}}$

$\displaystyle=\frac{{{\left({x}+{1}\right)}\sqrt{{{{\left({x}+{1}\right)}}^{{2}}-{1}}}}}{{2}}-\frac{{1}}{{2}}{\left({\ln}{\left|{\left({x}+{1}\right)}+\sqrt{{{{\left({x}+{1}\right)}}^{{2}}-{1}}}\right|}\right)}+{C}$

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Sunday, October 18, 2009

$\displaystyle\int\sqrt{{{{x}}^{{2}}+{2}{x}}}{\left.{d}{x}\right.}$ Evaluate the integral

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Sunday, October 18, 2009

Evaluate the integral

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Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle\int\sqrt{{{{x}}^{{2}}+{2}{x}}}{\left.{d}{x}\right.}$ Evaluate the integral

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?