Integration: show int sqrt(1+sqrt(x)) = (4/15)(1+sqrt(x))^(3

[chengeto]

In the following case use the given substitution to show that :

\(\displaystyle \int\sqrt(1+\sqrt x)=\frac{4}{15}(1+\sqrt x)^\frac{3}{2}(3\sqrt x-2)+c\)

\(\displaystyle u^2=1+\sqrt x\)


\(\displaystyle u=\sqrt(1+\sqrt x)\)


My attempt to solution:

\(\displaystyle u=\sqrt(1+\sqrt x)\)


\(\displaystyle du=\frac{1}{2\sqrt(1+\sqrt x )}.\frac{1}{2\sqrt x} dx\)


\(\displaystyle du=\frac{1}{4\sqrt(1+\sqrt x)(\sqrt x)}dx\)


I get stuck here l do not know how to substitute for du

 

[daon]

Re: Integration

\(\displaystyle u = \sqrt{x}+1\)

\(\displaystyle x = u^2 - 2u + 1\)

\(\displaystyle dx = (2u-2)du\)

\(\displaystyle \int \sqrt{\sqrt{x}+1}dx = \int \sqrt{u}(2u-2)du = \int u^{\frac{1}{2}}(2u)du - \int u^{\frac{1}{2}}(2)du\)

\(\displaystyle 2\int u^{\frac{3}{2}}du - 2\int u^{\frac{1}{2}}du\)


edit: whoops. looks like the directions want you to take the derivative of the given solution and show it is equal to your integrand.

 

[chengeto]

Re: Integration

\(\displaystyle u = \sqrt{x}+1\)

\(\displaystyle x = u^2 - 2u + 1\)

\(\displaystyle dx = (2u-2)du\)

\(\displaystyle \int \sqrt{\sqrt{x}+1}dx = \int \sqrt{u}(2u-2)du = \int u^{\frac{1}{2}}(2u)du - \int u^{\frac{1}{2}}(2)du\)

\(\displaystyle 2\int u^{\frac{3}{2}}du - 2\int u^{\frac{1}{2}}du\)


edit: whoops. looks like the directions want you to take the derivative of the given solution and show it is equal to your integrand.

Thanks for the much needed help

 

[Deleted member 4993]

Re: Integration

In the following case use the given substitution to show that :

\(\displaystyle \int\sqrt(1+\sqrt x)=\frac{4}{15}(1+\sqrt x)^\frac{3}{2}(3\sqrt x-2)+c\)

\(\displaystyle u^2=1+\sqrt x\)


\(\displaystyle u=\sqrt(1+\sqrt x)\)


My attempt to solution:

\(\displaystyle u=\sqrt(1+\sqrt x)\)


\(\displaystyle du=\frac{1}{2\sqrt(1+\sqrt x )}.\frac{1}{2\sqrt x} dx\)


\(\displaystyle du=\frac{dx}{4\sqrt(1+\sqrt x)(\sqrt x)}\)

\(\displaystyle dx \, = \, du \cdot 4u(u^2 - 1)\)

\(\displaystyle \int\sqrt(1+\sqrt x)dx\)

\(\displaystyle = \, \int u\cdot 4u(u^2 - 1)du\)

I get stuck here l do not know how to substitute for du

 

[chengeto]

Re: Integration

\(\displaystyle u = \sqrt{x}+1\)

\(\displaystyle x = u^2 - 2u + 1\)

\(\displaystyle dx = (2u-2)du\)

\(\displaystyle \int \sqrt{\sqrt{x}+1}dx = \int \sqrt{u}(2u-2)du = \int u^{\frac{1}{2}}(2u)du - \int u^{\frac{1}{2}}(2)du\)

\(\displaystyle 2\int u^{\frac{3}{2}}du - 2\int u^{\frac{1}{2}}du\)


edit: whoops. looks like the directions want you to take the derivative of the given solution and show it is equal to your integrand.

Daon :

\(\displaystyle u = \sqrt(1+\sqrt x)\)

 

[chengeto]

Re: Integration

In the following case use the given substitution to show that :

\(\displaystyle \int\sqrt(1+\sqrt x)=\frac{4}{15}(1+\sqrt x)^\frac{3}{2}(3\sqrt x-2)+c\)

\(\displaystyle u^2=1+\sqrt x\)


\(\displaystyle u=\sqrt(1+\sqrt x)\)


My attempt to solution:

\(\displaystyle u=\sqrt(1+\sqrt x)\)


\(\displaystyle du=\frac{1}{2\sqrt(1+\sqrt x )}.\frac{1}{2\sqrt x} dx\)


\(\displaystyle du=\frac{dx}{4\sqrt(1+\sqrt x)(\sqrt x)}\)

\(\displaystyle dx \, = \, du \cdot 4u(u^2 - 1)\)

\(\displaystyle \int\sqrt(1+\sqrt x)dx\)

\(\displaystyle = \, \int u\cdot 4u(u^2 - 1)du\)

I get stuck here l do not know how to substitute for du

\(\displaystyle \frac{4}{15}[3(\sqrt(1+\sqrt x)^5-5(\sqrt(1+\sqrt x)^3]=\frac{4}{15}(1+\sqrt x)^\frac{3}{2}(3\sqrt x-2)\)

What l don't understand is how do they end up with :

\(\displaystyle \frac{4}{15}(1+\sqrt x)^\frac{3}{2}(3\sqrt x-2)\)

 

[soroban]

Hello, chengeto!

\(\displaystyle \text{In the following, use: }\:u \:=\:\sqrt{1 + \sqrt{x}}\,\text{ to show that:}\)

. . \(\displaystyle \int\sqrt{1+\sqrt x}\,dx \;=\tfrac{4}{15}(1+\sqrt x)^{\frac{3}{2}}(3\sqrt x-2)+C\)

\(\displaystyle \text{We have: }\:u \:=\:\sqrt{1+\sqrt{x}} \quad\Rightarrow\quad u^2 \:=\:1 + \sqrt{x} \quad\Rightarrow\quad \sqrt{x} \:=\:u^2-1\)

. . \(\displaystyle x \:=\u^2-1)^2 \quad\Rightarrow\quad dx \;=\;2(u^2-1)(2u\,du) \;=\;4u(u^2-1)\,du\)


\(\displaystyle \text{Substitute: }\:\int u\cdot4u(u^2-1)\,du \;\;=\;\;4\int u^2(u^2-1)\,du\)

. . . . . . \(\displaystyle =\;\;4\int(u^4 - u^2)\,du \;\;=\;\;4\left(\frac{1}{5}u^5 - \frac{1}{3}u^3\right) + C\)

\(\displaystyle \text{Factor: }\:\frac{4}{15}u^3\left(3u^2 - 5\right) + C\)


\(\displaystyle \text{Back-substitute: }\:\frac{4}{15}\left(\sqrt{1+\sqrt{x}}\right)^3\bigg[3\left(1 + \sqrt{x}\right) - 5\bigg] + C\)

. . . . . . . . . . \(\displaystyle =\;\frac{4}{15}\left(1 + \sqrt{x}\right)^{\frac{3}{2}}\left(3\sqrt{x} - 2\right) + C\)

 

[chengeto]

Hello, chengeto!

\(\displaystyle \text{In the following, use: }\:u \:=\:\sqrt{1 + \sqrt{x}}\,\text{ to show that:}\)

. . \(\displaystyle \int\sqrt{1+\sqrt x}\,dx \;=\tfrac{4}{15}(1+\sqrt x)^{\frac{3}{2}}(3\sqrt x-2)+C\)

\(\displaystyle \text{We have: }\:u \:=\:\sqrt{1+\sqrt{x}} \quad\Rightarrow\quad u^2 \:=\:1 + \sqrt{x} \quad\Rightarrow\quad \sqrt{x} \:=\:u^2-1\)

. . \(\displaystyle x \:=\u^2-1)^2 \quad\Rightarrow\quad dx \;=\;2(u^2-1)(2u\,du) \;=\;4u(u^2-1)\,du\)


\(\displaystyle \text{Substitute: }\:\int u\cdot4u(u^2-1)\,du \;\;=\;\;4\int u^2(u^2-1)\,du\)

. . . . . . \(\displaystyle =\;\;4\int(u^4 - u^2)\,du \;\;=\;\;4\left(\frac{1}{5}u^5 - \frac{1}{3}u^3\right) + C\)

\(\displaystyle \text{Factor: }\:\frac{4}{15}u^3\left(3u^2 - 5\right) + C\)


\(\displaystyle \text{Back-substitute: }\:\frac{4}{15}\left(\sqrt{1+\sqrt{x}}\right)^3\bigg[3\left(1 + \sqrt{x}\right) - 5\bigg] + C\)

. . . . . . . . . . \(\displaystyle =\;\frac{4}{15}\left(1 + \sqrt{x}\right)^{\frac{3}{2}}\left(3\sqrt{x} - 2\right) + C\)

Thanks for the help. It seems as if l might need to work a little bit on my algebra