Solving Integrals with Substitution

[Yankel]

Hello,

I need some help solving this integral,

\[\int \frac{\sqrt{x}}{\sqrt{x}+1}dx...u=\sqrt{x}+1\][\tex]

After I make the substitution I get stuck a little bit, because I can't get rid of dx.

and also this one, same principle,

\[\int x^{3}\cdot \sqrt{7+3x}\cdot dx ...u=7+3x\]

how do I get rid of x^3 ?

thanks

 

[MarkFL]

Your substitution is a good one. If \(\displaystyle u=\sqrt{x}+1\), then can you express $dx$ in terms of $u$ only? You should find:

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

Now, from this, you may write:

\(\displaystyle dx=2\sqrt{x}\,dx\)

Can you rewrite \(\displaystyle \sqrt{x}\) in terms of $u$ using your original substitution?

 

[soroban]

Hello, Yankel!

$$\int \frac{\sqrt{x}}{\sqrt{x}+1}dx$$

Let $$u = \sqrt{x} \quad\Rightarrow\quad x = u^2 \quad\Rightarrow\quad dx = 2u\,du$$

Substitute: .$$\int \frac{u}{u+1}(2u\,du) \;=\;2\int\frac{u^2}{u+1}du$$

Long division: .$$2\int \left(u - 1 + \frac{1}{u+1}\right)\,du$$

. . . . $$=\;2\left(\tfrac{1}{2}u^2 - u + \ln|u+1|\right) + C$$

. . . . $$=\;u^2 - 2u + 2\ln|u+1| + C$$Back-substitute: .$$x - 2\sqrt{x} + 2\ln(\sqrt{x}+1) + C$$

 

[Prove It]

Hello,

I need some help solving this integral,

\[\int \frac{\sqrt{x}}{\sqrt{x}+1}dx...u=\sqrt{x}+1\][\tex]

After I make the substitution I get stuck a little bit, because I can't get rid of dx.

and also this one, same principle,

\[\int x^{3}\cdot \sqrt{7+3x}\cdot dx ...u=7+3x\]

how do I get rid of x^3 ?

thanks

\(\displaystyle \displaystyle \begin{align*} \int{\frac{\sqrt{x}}{\sqrt{x} + 1}\,dx} &= \int{\frac{\left( \sqrt{x} \right) ^2 }{\sqrt{x} \left( \sqrt{x} + 1 \right) }\,dx} \\ &= 2\int{ \frac{ \left( \sqrt{x} \right) ^2 }{\sqrt{x} + 1} \cdot \frac{1}{2\sqrt{x}}\,dx} \end{align*}\)

Now following your suggested substitution, let \(\displaystyle \displaystyle \begin{align*} u = \sqrt{x} + 1 \implies du = \frac{1}{2\sqrt{x}}\,dx \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} 2\int{ \frac{\left( \sqrt{x} \right) ^2}{\sqrt{x} + 1} \cdot \frac{1}{2\sqrt{x}}\,dx} &= 2\int{ \frac{ \left( u - 1 \right) ^2 }{ u } \, du} \\ &= 2\int{ \frac{u^2 - 2u + 1}{u} \, du } \\ &= 2\int{u - 2 + \frac{1}{u}\,du} \end{align*}\)

which can now be integrated.As for your second problem, \(\displaystyle \displaystyle \begin{align*} u = 7 + 3x \implies du = 3\,dx \end{align*}\) is a good start...

\(\displaystyle \displaystyle \begin{align*} \int{ x^3\,\sqrt{7 + 3x} \, dx} &= \frac{1}{3} \int{ 3x^3\,\sqrt{7 + 3x}\,dx} \\ &= \frac{1}{3} \int{ \left( \frac{u - 7 }{3} \right) ^3 \, \sqrt{u}\,du} \\ &= \frac{1}{3} \int{ \left( \frac{u^3 - 21u^2 + 147u - 343}{27} \right) u^{\frac{1}{2}} \,du} \\ &= \frac{1}{81} \int{ u^{\frac{7}{2}} - 21u^{\frac{5}{2}} + 147u^{\frac{3}{2}} - 343u^{\frac{1}{2}}\,du} \end{align*}\)

which can now be integrated.

 

[paulmdrdo1]

this is how would i solve the problem

\begin{align*}\displaystyle \int\frac{\sqrt{x}}{\sqrt{x}+1}dx\\& let\,\,u\,=\sqrt{x}+1\\&\sqrt{x}\,=\,u-1\\& du\,=\,\frac{1}{2}x^{-\frac{1}{2}}dx\\& dx\,=\,\frac{2}{x^{-\frac{1}{2}}}du\\& rewrite\,\,dx\,as\,\,2\sqrt{x}\,du\\ then\,,\\\end{align*}
\begin{align*}\displaystyle \int\frac{\sqrt{x}}{\sqrt{x}+1}dx=\int\frac{u-1}{u}2(u-1)du\\ we\,\,now\,\, have\,\,\int\frac{2u^2-4u+2}{u}du\,\,(integrate\,\,the\,\,terms\,\,individualy)\\\end{align*}

\begin{align*}\displaystyle2\int udu-4\int du+2\int\frac{1}{u}du\\&2u^2-4u+2\ln|u|+c =(\sqrt{x}+1)^2-4\sqrt{x}+2\ln|\sqrt{x}|+c \\&expand(\sqrt{x}+1)^2 \\&=\,\,x-\sqrt{x}+2\ln|\sqrt{x}+1|+C\end{align*}

Hope this would help.:)