Question

Evaluate the following integrals
${\int }_2^3\sqrt{(x^2+6x-7)}dx$

Answer 1

${\int }_2^3\sqrt{x^2+6x-7}dx$

Adding and subtracting 9 under the square root, we get

$\Rightarrow {\int }_2^3\sqrt{x^2+6x+9-9-7}dx$

$\Rightarrow {\int }_2^3\sqrt{(x+3{)}^2-16}dx$

Let $u=x+3$

Then, $du=dx$

Upper limit: at $x=3$, $u=x+3=3+3=6$

Lower limit: at $x=2$, $u=x+3=2+3=5$

$\Rightarrow {\int }_5^6\sqrt{u^2-16}du$

Let, $I=\int \sqrt{u^2-16}du...\left(i\right)$

Using integration by parts,

$I=\int \sqrt{u^2-16}.1du$

$\Rightarrow I=\sqrt{u^2-16}\int 1du-\int [\frac{d}{du}\sqrt{u^2-16}\times \int 1du]du$

$\Rightarrow I=u\sqrt{u^2-16}-\int \frac{1}{2\sqrt{u^2-16}}.2u\times udu$

$\Rightarrow I=u\sqrt{u^2-16}-\int \frac{u^2}{\sqrt{u^2-16}}du$

$\Rightarrow I=u\sqrt{u^2-16}-\int \frac{u^2-16+16}{\sqrt{u^2-16}}du$

$\Rightarrow I=u\sqrt{u^2-16}-\int \frac{u^2-16}{\sqrt{u^2-16}}du-\int \frac{16}{\sqrt{u^2-16}}du$

$\Rightarrow I=u\sqrt{u^2-16}-\int \sqrt{u^2-16}du-\int \frac{16}{\sqrt{u^2-16}}du$

$\Rightarrow I=u\sqrt{u^2-16}-I-\int \frac{16}{\sqrt{u^2-16}}du$

$\Rightarrow 2I=u\sqrt{u^2-16}-\int \frac{16}{\sqrt{u^2-16}}du$

$\Rightarrow I=\frac{1}{2}u\sqrt{u^2-16}-8\ln (u+\sqrt{u^2-16})$

Now substituting the upper and lower limits of integration we get,

${\int }_5^6\sqrt{u^2-16}du=|\frac{1}{2}u\sqrt{u^2-16}-8\ln (u+\sqrt{u^2-16}){|}_5^6$

$=\frac{1}{2}(6\sqrt{6^2-16}-5\sqrt{5^2-16})-8\left(\ln \right(6+\sqrt{6^2-16})-\ln (5+\sqrt{5^2-16})$

$=\frac{1}{2}(6\sqrt{20}-5\sqrt{9})-8\left(\ln \right(6+\sqrt{20})-\ln (5+\sqrt{9}\left)\right)$

$=\frac{1}{2}(12\sqrt{5}-15)-8\ln \frac{6+2\sqrt{5}}{8}$

$=\frac{1}{2}(12\sqrt{5}-15)-8\ln \frac{3+\sqrt{5}}{4}$