Solving Integrals with Substitution

In summary, Yankel found that in two simple problems, substitution and long division, he could solve the integral by integrating the terms individually.
  • #1
Yankel
395
0
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
 
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  • #2
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?
 
  • #3
Hello, Yankel!

[tex]\int \frac{\sqrt{x}}{\sqrt{x}+1}dx [/tex]

Let [tex]u = \sqrt{x} \quad\Rightarrow\quad x = u^2 \quad\Rightarrow\quad dx = 2u\,du[/tex]

Substitute: .[tex]\int \frac{u}{u+1}(2u\,du) \;=\;2\int\frac{u^2}{u+1}du [/tex]

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

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

. . . . [tex]=\;u^2 - 2u + 2\ln|u+1| + C[/tex]Back-substitute: .[tex]x - 2\sqrt{x} + 2\ln(\sqrt{x}+1) + C[/tex]
 
  • #4
Yankel said:
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.
 
  • #5
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.:)
 

FAQ: Solving Integrals with Substitution

What is integration with substitution?

Integration with substitution is a technique used in calculus to solve indefinite integrals. It involves substituting a variable in the integral with another variable or expression in order to simplify the integral and make it easier to solve.

When should I use integration with substitution?

Integration with substitution is useful when the integrand (the expression being integrated) involves a function that is composed of another function. It is also helpful when the integral involves a complicated expression that can be simplified by substituting a variable.

How do I choose the substitution variable?

The substitution variable is typically chosen to be a part of the expression inside the integral. It should also be chosen such that it simplifies the expression and makes the integral easier to solve. Common substitution variables include u, x, t, and θ.

What is the general process for integration with substitution?

The general process for integration with substitution is as follows:
1. Identify the substitution variable and its derivative.
2. Substitute the variable and its derivative into the integral.
3. Simplify the integral using the substitution.
4. Integrate the simplified integral.
5. Substitute the original variable back into the result.

Are there any common mistakes to avoid when using integration with substitution?

One common mistake to avoid is forgetting to substitute the derivative of the substitution variable when simplifying the integral. Another mistake is choosing a substitution variable that does not simplify the integral or makes it more complicated. It is also important to remember to substitute the original variable back into the final result.

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