New Notes Blog: `int (4x-2/(2x+3)^2) dx` Find the indefinite integral

Sunday, 16 March 2014

$\displaystyle\int{\left({4}{x}-\frac{{2}}{{{\left({2}{x}+{3}\right)}}^{{2}}}\right)}{\left.{d}{x}\right.}$ Find the indefinite integral

$\displaystyle\int{\left({4}{x}-\frac{{2}}{{{\left({2}{x}+{3}\right)}}^{{2}}}\right)}{\left.{d}{x}\right.}$

To solve, express it as difference of two integrals.

$\displaystyle=\int{4}{x}{\left.{d}{x}\right.}-\int\frac{{2}}{{{\left({2}{x}+{3}\right)}}^{{2}}}{\left.{d}{x}\right.}$

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

$\displaystyle=\int{4}{x}{\left.{d}{x}\right.}-\int{2}{{\left({2}{x}+{3}\right)}}^{{-{2}}}{\left.{d}{x}\right.}$

For the second integral, apply the u-substitution method.

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

$\displaystyle{d}{u}={2}{\left.{d}{x}\right.}$

Expressing the second integral in terms of u variable, it becomes:

$\displaystyle=\int{4}{x}{\left.{d}{x}\right.}-\int{{\left({2}{x}+{3}\right)}}^{{-{2}}}\cdot{2}{\left.{d}{x}\right.}$

$\displaystyle=\int{4}{x}{\left.{d}{x}\right.}-\int{{u}}^{{-{2}}}{d}{u}$

For both...

$\displaystyle\int{\left({4}{x}-\frac{{2}}{{{\left({2}{x}+{3}\right)}}^{{2}}}\right)}{\left.{d}{x}\right.}$

To solve, express it as difference of two integrals.

$\displaystyle=\int{4}{x}{\left.{d}{x}\right.}-\int\frac{{2}}{{{\left({2}{x}+{3}\right)}}^{{2}}}{\left.{d}{x}\right.}$

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

$\displaystyle=\int{4}{x}{\left.{d}{x}\right.}-\int{2}{{\left({2}{x}+{3}\right)}}^{{-{2}}}{\left.{d}{x}\right.}$

For the second integral, apply the u-substitution method.

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

$\displaystyle{d}{u}={2}{\left.{d}{x}\right.}$

Expressing the second integral in terms of u variable, it becomes:

$\displaystyle=\int{4}{x}{\left.{d}{x}\right.}-\int{{\left({2}{x}+{3}\right)}}^{{-{2}}}\cdot{2}{\left.{d}{x}\right.}$

$\displaystyle=\int{4}{x}{\left.{d}{x}\right.}-\int{{u}}^{{-{2}}}{d}{u}$

For both integrals, apply the formula $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

$\displaystyle=\frac{{{4}{{x}}^{{2}}}}{{2}}-\frac{{{u}}^{{-{1}}}}{{-{1}}}+{C}$

$\displaystyle={2}{{x}}^{{2}}+{{u}}^{{-{1}}}+{C}$

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

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

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

Therefore, $\displaystyle\int{\left({4}{x}-\frac{{2}}{{{\left({2}{x}+{3}\right)}}^{{2}}}\right)}{\left.{d}{x}\right.}={2}{{x}}^{{2}}+\frac{{1}}{{{2}{x}+{3}}}+{C}$ .

New Notes Blog

Sunday, 16 March 2014

$\displaystyle\int{\left({4}{x}-\frac{{2}}{{{\left({2}{x}+{3}\right)}}^{{2}}}\right)}{\left.{d}{x}\right.}$ Find the indefinite integral

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

Sunday, 16 March 2014

Find the indefinite integral

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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?

$\displaystyle\int{\left({4}{x}-\frac{{2}}{{{\left({2}{x}+{3}\right)}}^{{2}}}\right)}{\left.{d}{x}\right.}$ 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?