Find the integral: − <msqrt> t 2 </mrow> + t

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

${\displaystyle -\int \sqrt{t^2+t-5}dt}$

Complete the square.

${\displaystyle -\int \sqrt{{(t+\frac{1}{2})}^2-\frac{21}{4}}dt}$

Let ${\displaystyle u=t+\frac{1}{2}}$. Then ${\displaystyle du=dt}$. Rewrite using ${\displaystyle u}$ and ${\displaystyle d}$${\displaystyle u}$.

${\displaystyle -\int \sqrt{u^2-\frac{21}{4}}du}$

To write ${\displaystyle u^2}$ as a fraction with a common denominator, multiply by ${\displaystyle \frac{4}{4}}$.

${\displaystyle -\int \sqrt{u^2\cdot \frac{4}{4}-\frac{21}{4}}du}$

Simplify terms.

${\displaystyle -\int \sqrt{\frac{u^2\cdot 4-21}{4}}du}$

Move ${\displaystyle 4}$ to the left of ${\displaystyle u^2}$.

${\displaystyle -\int \sqrt{\frac{4u^2-21}{4}}du}$

Rewrite ${\displaystyle \frac{4u^2-21}{4}}$ as ${\displaystyle {\left(\frac{1}{2}\right)}^2(4u^2-21)}$.

${\displaystyle -\int \sqrt{{\left(\frac{1}{2}\right)}^2(4u^2-21)}du}$

Simplify terms.

${\displaystyle -\int \frac{\sqrt{4u^2-21}}{2}du}$

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

${\displaystyle -(\frac{1}{2}\int \sqrt{4u^2-21}du)}$

Let ${\displaystyle u=\frac{\sqrt{21}}{2}\sec \left(t_1\right)}$, where ${\displaystyle -\frac{\pi }{2}\le t_1\le \frac{\pi }{2}}$. Then ${\displaystyle du=\frac{\sqrt{21}\sec \left(t_1\right)\tan \left(t_1\right)}{2}d{t}_1}$. Note that since ${\displaystyle -\frac{\pi }{2}\le t_1\le \frac{\pi }{2}}$, ${\displaystyle \frac{\sqrt{21}\sec \left(t_1\right)\tan \left(t_1\right)}{2}}$ is positive.

${\displaystyle -\frac{1}{2}\int \sqrt{4{\left(\frac{\sqrt{21}}{2}\sec \left(t_1\right)\right)}^2-21}\frac{\sqrt{21}\sec \left(t_1\right)\tan \left(t_1\right)}{2}d{t}_1}$

Simplify terms.

${\displaystyle -\frac{1}{2}\int \frac{21\sec \left(t_1\right){\tan }^2\left(t_1\right)}{2}d{t}_1}$

Since ${\displaystyle \frac{21}{2}}$ is constant with respect to ${\displaystyle t_1}$, move ${\displaystyle \frac{21}{2}}$ out of the integral.

${\displaystyle -\frac{1}{2}(\frac{21}{2}\int \sec \left(t_1\right){\tan }^2\left(t_1\right)d{t}_1)}$

Simplify.

${\displaystyle -\frac{21}{4}\int \sec \left(t_1\right){\tan }^2\left(t_1\right)d{t}_1}$

Raise ${\displaystyle \sec \left(t_1\right)}$ to the power of ${\displaystyle 1}$.

${\displaystyle -\frac{21}{4}\int {\sec }^1\left(t_1\right){\tan }^2\left(t_1\right)d{t}_1}$

Using the Pythagorean Identity, rewrite ${\displaystyle {\tan }^2\left(t_1\right)}$ as ${\displaystyle -1+{\sec }^2\left(t_1\right)}$.

${\displaystyle -\frac{21}{4}\int {\sec }^1\left(t_1\right)(-1+{\sec }^2\left(t_1\right))d{t}_1}$

Simplify terms.

${\displaystyle -\frac{21}{4}\int -\sec \left(t_1\right)+{\sec }^3\left(t_1\right)d{t}_1}$

Split the single integral into multiple integrals.

${\displaystyle -\frac{21}{4}(\int -\sec \left(t_1\right)d{t}_1+\int {\sec }^3\left(t_1\right)d{t}_1)}$

Since ${\displaystyle -1}$ is constant with respect to ${\displaystyle t_1}$, move ${\displaystyle -1}$ out of the integral.

${\displaystyle -\frac{21}{4}(-\int \sec \left(t_1\right)d{t}_1+\int {\sec }^3\left(t_1\right)d{t}_1)}$

The integral of ${\displaystyle \sec \left(t_1\right)}$ with respect to ${\displaystyle t_1}$ is ${\displaystyle \ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\int {\sec }^3\left(t_1\right)d{t}_1)}$

Factor ${\displaystyle \sec \left(t_1\right)}$ out of ${\displaystyle {\sec }^3\left(t_1\right)}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\int \sec \left(t_1\right){\sec }^2\left(t_1\right)d{t}_1)}$

Integrate by parts using the formula ${\displaystyle \int udv=uv-\int vdu}$, where ${\displaystyle u=\sec \left(t_1\right)}$ and ${\displaystyle dv={\sec }^2\left(t_1\right)}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int \tan \left(t_1\right)\left(\sec \left(t_1\right)\tan \left(t_1\right)\right)d{t}_1)}$

Raise ${\displaystyle \tan \left(t_1\right)}$ to the power of ${\displaystyle 1}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int {\tan }^1\left(t_1\right)\tan \left(t_1\right)\sec \left(t_1\right)d{t}_1)}$

Raise ${\displaystyle \tan \left(t_1\right)}$ to the power of ${\displaystyle 1}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int {\tan }^1\left(t_1\right){\tan }^1\left(t_1\right)\sec \left(t_1\right)d{t}_1)}$

Use the power rule ${\displaystyle a^m{a}^n=a^{m+n}}$ to combine exponents.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int {\tan \left(t_1\right)}^{1+1}\sec \left(t_1\right)d{t}_1)}$

Simplify the expression.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int \sec \left(t_1\right){\tan }^2\left(t_1\right)d{t}_1)}$

Using the Pythagorean Identity, rewrite ${\displaystyle {\tan }^2\left(t_1\right)}$ as ${\displaystyle -1+{\sec }^2\left(t_1\right)}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int \sec \left(t_1\right)(-1+{\sec }^2\left(t_1\right))d{t}_1)}$

Simplify by multiplying through.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int -1\cdot \sec \left(t_1\right)+\sec \left(t_1\right)\left(\sec \left(t_1\right)\sec \left(t_1\right)\right)d{t}_1)}$

Raise ${\displaystyle \sec \left(t_1\right)}$ to the power of ${\displaystyle 1}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int -1\sec \left(t_1\right)+{\sec }^1\left(t_1\right)\sec \left(t_1\right)\sec \left(t_1\right)d{t}_1)}$

Raise ${\displaystyle \sec \left(t_1\right)}$ to the power of ${\displaystyle 1}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int -1\sec \left(t_1\right)+{\sec }^1\left(t_1\right){\sec }^1\left(t_1\right)\sec \left(t_1\right)d{t}_1)}$

Use the power rule ${\displaystyle a^m{a}^n=a^{m+n}}$ to combine exponents.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int -1\sec \left(t_1\right)+{\sec \left(t_1\right)}^{1+1}\sec \left(t_1\right)d{t}_1)}$

Add ${\displaystyle 1}$ and ${\displaystyle 1}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int -1\sec \left(t_1\right)+{\sec }^2\left(t_1\right)\sec \left(t_1\right)d{t}_1)}$

Raise ${\displaystyle \sec \left(t_1\right)}$ to the power of ${\displaystyle 1}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int -1\sec \left(t_1\right)+{\sec }^2\left(t_1\right){\sec }^1\left(t_1\right)d{t}_1)}$

Use the power rule ${\displaystyle a^m{a}^n=a^{m+n}}$ to combine exponents.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int -1\sec \left(t_1\right)+{\sec \left(t_1\right)}^{2+1}d{t}_1)}$

Add ${\displaystyle 2}$ and ${\displaystyle 1}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-\int -1\sec \left(t_1\right)+{\sec }^3\left(t_1\right)d{t}_1)}$

Split the single integral into multiple integrals.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-(\int -1\sec \left(t_1\right)d{t}_1+\int {\sec }^3\left(t_1\right)d{t}_1))}$

Since ${\displaystyle -1}$ is constant with respect to ${\displaystyle t_1}$, move ${\displaystyle -1}$ out of the integral.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-(-\int \sec \left(t_1\right)d{t}_1+\int {\sec }^3\left(t_1\right)d{t}_1))}$

The integral of ${\displaystyle \sec \left(t_1\right)}$ with respect to ${\displaystyle t_1}$ is ${\displaystyle \ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)-(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\int {\sec }^3\left(t_1\right)d{t}_1))}$

Simplify by multiplying through.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\sec \left(t_1\right)\tan \left(t_1\right)+1(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)-\int {\sec }^3\left(t_1\right)d{t}_1)}$

Solving for ${\displaystyle \int {\sec }^3\left(t_1\right)d{t}_1}$, we find that ${\displaystyle \int {\sec }^3\left(t_1\right)d{t}_1}$ ${\displaystyle \frac{\sec \left(t_1\right)\tan \left(t_1\right)+1(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)}{2}}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\frac{\sec \left(t_1\right)\tan \left(t_1\right)+1(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)}{2}+C)}$

Multiply ${\displaystyle \ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C}$ by ${\displaystyle 1}$.

${\displaystyle -\frac{21}{4}(-(\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C)+\frac{\sec \left(t_1\right)\tan \left(t_1\right)+\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)+C}{2}+C)}$

Simplify.

${\displaystyle -\frac{21}{4}\cdot \frac{-\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)\cdot 2+\sec \left(t_1\right)\tan \left(t_1\right)+\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right)}{2}+C}$

Simplify.

${\displaystyle -\frac{21(\sec \left(t_1\right)\tan \left(t_1\right)-\ln \left(|\sec \left(t_1\right)+\tan \left(t_1\right)|\right))}{8}+C}$

Substitute back in for each integration substitution variable.

${\displaystyle -\frac{21(\sec \left(\mathrm{arcsec}\left(\frac{2(t+\frac{1}{2})}{\sqrt{21}}\right)\right)\tan \left(\mathrm{arcsec}\left(\frac{2(t+\frac{1}{2})}{\sqrt{21}}\right)\right)-\ln \left(|\sec \left(\mathrm{arcsec}\left(\frac{2(t+\frac{1}{2})}{\sqrt{21}}\right)\right)+\tan \left(\mathrm{arcsec}\left(\frac{2(t+\frac{1}{2})}{\sqrt{21}}\right)\right)|\right))}{8}+C}$

Simplify.

${\displaystyle -\frac{21(\sec \left(\mathrm{arcsec}\left(\frac{2t+1}{\sqrt{21}}\right)\right)\tan \left(\mathrm{arcsec}\left(\frac{2t+1}{\sqrt{21}}\right)\right)-\ln \left(|\sec \left(\mathrm{arcsec}\left(\frac{2t+1}{\sqrt{21}}\right)\right)+\tan \left(\mathrm{arcsec}\left(\frac{2t+1}{\sqrt{21}}\right)\right)|\right))}{8}+C}$

Reorder terms.

${\displaystyle -\frac{21}{8}(\sec \left(\mathrm{arcsec}\left(\frac{1}{\sqrt{21}}(2t+1)\right)\right)\tan \left(\mathrm{arcsec}\left(\frac{1}{\sqrt{21}}(2t+1)\right)\right)-\ln \left(|\sec \left(\mathrm{arcsec}\left(\frac{1}{\sqrt{21}}(2t+1)\right)\right)+\tan \left(\mathrm{arcsec}\left(\frac{1}{\sqrt{21}}(2t+1)\right)\right)|\right))+C}$