New Notes Blog: `int (x+1) / sqrt(3x^2+6x) dx` Find the indefinite integral

Wednesday, 25 December 2013

$\displaystyle\int\frac{{{x}+{1}}}{\sqrt{{{3}{{x}}^{{2}}+{6}{x}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

$\displaystyle\int\frac{{{x}+{1}}}{\sqrt{{{3}{{x}}^{{2}}+{6}{x}}}}{\left.{d}{x}\right.}$

To solve, apply u-substitution method.

$\displaystyle{u}={3}{{x}}^{{2}}+{6}{x}$

$\displaystyle{d}{u}={\left({6}{x}+{6}\right)}{\left.{d}{x}\right.}$

$\displaystyle{d}{u}={6}{\left({x}+{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle\frac{{1}}{{6}}{d}{u}={\left({x}+{1}\right)}{\left.{d}{x}\right.}$

Expressing the integral in terms of u, it becomes

$\displaystyle=\int\frac{{1}}{\sqrt{{{3}{{x}}^{{2}}+{6}{x}}}}\cdot{\left({x}+{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{\sqrt{{u}}}\cdot\frac{{1}}{{6}}{d}{u}$

$\displaystyle=\frac{{1}}{{6}}\int\frac{{1}}{\sqrt{{u}}}{d}{u}$

Then, convert the radical to exponent form.

$\displaystyle=\frac{{1}}{{6}}\int\frac{{1}}{{{u}}^{{\frac{{1}}{{2}}}}}{d}{u}$

Also, apply the negative exponent rule $\displaystyle{{a}}^{{-{m}}}=\frac{{1}}{{{a}}^{{m}}}$ .

$\displaystyle=\frac{{1}}{{6}}\int{{u}}^{{-\frac{{1}}{{2}}}}\ldots$

$\displaystyle\int\frac{{{x}+{1}}}{\sqrt{{{3}{{x}}^{{2}}+{6}{x}}}}{\left.{d}{x}\right.}$

To solve, apply u-substitution method.

$\displaystyle{u}={3}{{x}}^{{2}}+{6}{x}$

$\displaystyle{d}{u}={\left({6}{x}+{6}\right)}{\left.{d}{x}\right.}$

$\displaystyle{d}{u}={6}{\left({x}+{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle\frac{{1}}{{6}}{d}{u}={\left({x}+{1}\right)}{\left.{d}{x}\right.}$

Expressing the integral in terms of u, it becomes

$\displaystyle=\int\frac{{1}}{\sqrt{{{3}{{x}}^{{2}}+{6}{x}}}}\cdot{\left({x}+{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{\sqrt{{u}}}\cdot\frac{{1}}{{6}}{d}{u}$

$\displaystyle=\frac{{1}}{{6}}\int\frac{{1}}{\sqrt{{u}}}{d}{u}$

Then, convert the radical to exponent form.

$\displaystyle=\frac{{1}}{{6}}\int\frac{{1}}{{{u}}^{{\frac{{1}}{{2}}}}}{d}{u}$

Also, apply the negative exponent rule $\displaystyle{{a}}^{{-{m}}}=\frac{{1}}{{{a}}^{{m}}}$ .

$\displaystyle=\frac{{1}}{{6}}\int{{u}}^{{-\frac{{1}}{{2}}}}{d}{u}$

To take the integral of this, apply the formula $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

$\displaystyle=\frac{{1}}{{6}}\cdot\frac{{{u}}^{{\frac{{1}}{{2}}}}}{{\frac{{1}}{{2}}}}+{C}$

$\displaystyle=\frac{{1}}{{6}}\cdot\frac{{{2}{{u}}^{{\frac{{1}}{{2}}}}}}{{1}}+{C}$

$\displaystyle=\frac{{{u}}^{{\frac{{1}}{{2}}}}}{{3}}+{C}$

$\displaystyle=\frac{\sqrt{{u}}}{{3}}+{C}$

And, substitute back $\displaystyle{u}={3}{{x}}^{{2}}+{6}{x}$ .

$\displaystyle=\frac{\sqrt{{{3}{{x}}^{{2}}+{6}{x}}}}{{3}}+{C}$

Therefore, $\displaystyle\int\frac{{{x}+{1}}}{\sqrt{{{3}{{x}}^{{2}}+{6}{x}}}}{\left.{d}{x}\right.}=\frac{\sqrt{{{3}{{x}}^{{2}}+{6}{x}}}}{{3}}+{C}$ .

New Notes Blog

Wednesday, 25 December 2013

$\displaystyle\int\frac{{{x}+{1}}}{\sqrt{{{3}{{x}}^{{2}}+{6}{x}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

No comments:

Post a Comment

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Wednesday, 25 December 2013

Find the indefinite integral

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle\int\frac{{{x}+{1}}}{\sqrt{{{3}{{x}}^{{2}}+{6}{x}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?