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This integral could not be completed using integration by parts. Mathway will use another method.

Complete the square.

${\displaystyle \int \sqrt{-3{(x-\frac{1}{6})}^2+\frac{49}{12}}dx}$

Let ${\displaystyle u_1=x-\frac{1}{6}}$. Then ${\displaystyle d{u}_1=dx}$. Rewrite using ${\displaystyle u_1}$ and ${\displaystyle d}$${\displaystyle u_1}$.

${\displaystyle \int \sqrt{-3{u_1}^2+\frac{49}{12}}d{u}_1}$

To write ${\displaystyle -3{u_1}^2}$ as a fraction with a common denominator, multiply by ${\displaystyle \frac{12}{12}}$.

${\displaystyle \int \sqrt{-3{u_1}^2\cdot \frac{12}{12}+\frac{49}{12}}d{u}_1}$

Simplify terms.

${\displaystyle \int \sqrt{\frac{-3{u_1}^2\cdot 12+49}{12}}d{u}_1}$

Simplify the numerator.

${\displaystyle \int \sqrt{\frac{(7+6u_1)(7-6u_1)}{12}}d{u}_1}$

Rewrite ${\displaystyle \frac{(7+6u_1)(7-6u_1)}{12}}$ as ${\displaystyle {\left(\frac{1}{2}\right)}^2\frac{(7+6u_1)(7-6u_1)}{3}}$.

${\displaystyle \int \sqrt{{\left(\frac{1}{2}\right)}^2\frac{(7+6u_1)(7-6u_1)}{3}}d{u}_1}$

Pull terms out from under the radical.

${\displaystyle \int \frac{1}{2}\sqrt{\frac{(7+6u_1)(7-6u_1)}{3}}d{u}_1}$

Rewrite ${\displaystyle \sqrt{\frac{(7+6u_1)(7-6u_1)}{3}}}$ as ${\displaystyle \frac{\sqrt{(7+6u_1)(7-6u_1)}}{\sqrt{3}}}$.

${\displaystyle \int \frac{1}{2}\cdot \frac{\sqrt{(7+6u_1)(7-6u_1)}}{\sqrt{3}}d{u}_1}$

Multiply ${\displaystyle \frac{\sqrt{(7+6u_1)(7-6u_1)}}{\sqrt{3}}}$ by ${\displaystyle \frac{\sqrt{3}}{\sqrt{3}}}$.

${\displaystyle \int \frac{1}{2}(\frac{\sqrt{(7+6u_1)(7-6u_1)}}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}})d{u}_1}$

Combine and simplify the denominator.

${\displaystyle \int \frac{1}{2}\cdot \frac{\sqrt{(7+6u_1)(7-6u_1)}\sqrt{3}}{3}d{u}_1}$

Combine using the product rule for radicals.

${\displaystyle \int \frac{1}{2}\cdot \frac{\sqrt{(7+6u_1)(7-6u_1)\cdot 3}}{3}d{u}_1}$

Multiply ${\displaystyle \frac{1}{2}\cdot \frac{\sqrt{(7+6u_1)(7-6u_1)\cdot 3}}{3}}$.

${\displaystyle \int \frac{\sqrt{(7+6u_1)(7-6u_1)\cdot 3}}{6}d{u}_1}$

Simplify by multiplying through.

${\displaystyle \int \frac{\sqrt{(21+18u_1)(7-6u_1)}}{6}d{u}_1}$

Expand ${\displaystyle (21+18u_1)(7-6u_1)}$ using the FOIL Method.

${\displaystyle \int \frac{\sqrt{21\cdot 7+21(-6u_1)+18u_1\cdot 7+18u_1(-6u_1)}}{6}d{u}_1}$

Simplify and combine like terms.

${\displaystyle \int \frac{\sqrt{147-108{u_1}^2}}{6}d{u}_1}$

Since ${\displaystyle \frac{1}{6}}$ is constant with respect to ${\displaystyle u_1}$, move ${\displaystyle \frac{1}{6}}$ out of the integral.

${\displaystyle \frac{1}{6}\int \sqrt{147-108{u_1}^2}d{u}_1}$

Let ${\displaystyle u_1=\frac{7}{6}\sin \left(t\right)}$, where ${\displaystyle -\frac{\pi }{2}\le t\le \frac{\pi }{2}}$. Then ${\displaystyle d{u}_1=\frac{7\cos \left(t\right)}{6}dt}$. Note that since ${\displaystyle -\frac{\pi }{2}\le t\le \frac{\pi }{2}}$, ${\displaystyle \frac{7\cos \left(t\right)}{6}}$ is positive.

${\displaystyle \frac{1}{6}\int \sqrt{{\left(7\sqrt{3}\right)}^2-108{\left(\frac{7}{6}\sin \left(t\right)\right)}^2}\frac{7\cos \left(t\right)}{6}dt}$

Simplify terms.

${\displaystyle \frac{1}{6}\int \frac{49{\cos }^2\left(t\right)\sqrt{3}}{6}dt}$

Since ${\displaystyle \frac{49\sqrt{3}}{6}}$ is constant with respect to ${\displaystyle t}$, move ${\displaystyle \frac{49\sqrt{3}}{6}}$ out of the integral.

${\displaystyle \frac{1}{6}(\frac{49\sqrt{3}}{6}\int {\cos }^2\left(t\right)dt)}$

Simplify.

${\displaystyle \frac{49\sqrt{3}}{36}\int {\cos }^2\left(t\right)dt}$

Use the half-angle formula to rewrite ${\displaystyle {\cos }^2\left(t\right)}$ as ${\displaystyle \frac{1+\cos \left(2t\right)}{2}}$.

${\displaystyle \frac{49\sqrt{3}}{36}\int \frac{1+\cos \left(2t\right)}{2}dt}$

Since ${\displaystyle \frac{1}{2}}$ is constant with respect to ${\displaystyle t}$, move ${\displaystyle \frac{1}{2}}$ out of the integral.

${\displaystyle \frac{49\sqrt{3}}{36}(\frac{1}{2}\int 1+\cos \left(2t\right)dt)}$

Simplify.

${\displaystyle \frac{49\sqrt{3}}{72}\int 1+\cos \left(2t\right)dt}$

Split the single integral into multiple integrals.

${\displaystyle \frac{49\sqrt{3}}{72}(\int dt+\int \cos \left(2t\right)dt)}$

Apply the constant rule.

${\displaystyle \frac{49\sqrt{3}}{72}(t+C+\int \cos \left(2t\right)dt)}$

Let ${\displaystyle u_2=2t}$. Then ${\displaystyle d{u}_2=2dt}$, so ${\displaystyle \frac{1}{2}d{u}_2=dt}$. Rewrite using ${\displaystyle u_2}$ and ${\displaystyle d}$${\displaystyle u_2}$.

${\displaystyle \frac{49\sqrt{3}}{72}(t+C+\int \cos \left(u_2\right)\frac{1}{2}d{u}_2)}$

Combine ${\displaystyle \cos \left(u_2\right)}$ and ${\displaystyle \frac{1}{2}}$.

${\displaystyle \frac{49\sqrt{3}}{72}(t+C+\int \frac{\cos \left(u_2\right)}{2}d{u}_2)}$

Since ${\displaystyle \frac{1}{2}}$ is constant with respect to ${\displaystyle u_2}$, move ${\displaystyle \frac{1}{2}}$ out of the integral.

${\displaystyle \frac{49\sqrt{3}}{72}(t+C+\frac{1}{2}\int \cos \left(u_2\right)d{u}_2)}$

The integral of ${\displaystyle \cos \left(u_2\right)}$ with respect to ${\displaystyle u_2}$ is ${\displaystyle \sin \left(u_2\right)}$.

${\displaystyle \frac{49\sqrt{3}}{72}(t+C+\frac{1}{2}(\sin \left(u_2\right)+C))}$

Simplify.

${\displaystyle \frac{49\sqrt{3}}{72}(t+\frac{1}{2}\sin \left(u_2\right))+C}$

Substitute back in for each integration substitution variable.

${\displaystyle \frac{49\sqrt{3}}{72}(\arcsin \left(\frac{6(x-\frac{1}{6})}{7}\right)+\frac{1}{2}\sin \left(2\arcsin \left(\frac{6(x-\frac{1}{6})}{7}\right)\right))+C}$

Simplify.

${\displaystyle \frac{49\sqrt{3}\arcsin \left(\frac{6x-1}{7}\right)}{72}+\frac{49\sqrt{3}\sin \left(2\arcsin \left(\frac{6x-1}{7}\right)\right)}{144}+C}$

Reorder terms.

$\frac{49\sqrt{3}}{72}\arcsin \left(\frac{{\displaystyle 6x-1}}{{\displaystyle 7}}\right)+\frac{49\sqrt{3}}{144}\sin \left(2\arcsin \left(\frac{6x-1}{7}\right)\right)+C$