Answer to Question #223111 in Calculus for Nama Glory

Question

Evaluate the following Integrals

3)  ʃ_-1¹ (x + 1/2) √(x²+x+1)dx

4)  ʃ (x+2) / √(x+3) dx

Answer 1

3)

$\int_{-1}^1 (x +\frac{1}{2}) \sqrt{(x^2+x+1)}dx$

Solution

$\int (x +\frac{1}{2}) \sqrt{(x^2+x+1)}dx$

Substitute. $u=x^2+x+1$

$\frac{du}{dx}=2x+1 \to dx=\frac{1}{2x+1}du:$

$=\frac{1}{2} \int \sqrt{u}du$

Now solving:

$=\int \sqrt{u}du$

Apply power rule

$=\int u^ndu=\frac{u^{n+1}}{n+1}$ with $n=\frac{1}{2}$

$=\frac{2u^{\frac{3}{2}}}{3}$

Plug in solved integrals

$=\frac{1}{2} \int \sqrt{u}du$

$=\frac{u^{\frac{3}{2}}}{3}$

Undo substitution $u=x^2+x+1:$

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

Inserting the bounds

$=[\frac{(1^2+1+1)^{\frac{3}{2}}}{3}+C] -[\frac{((-1)^2+(-1)+1)^{\frac{3}{2}}}{3}+C]$

$\sqrt{3}-\frac{1}{3}=1.398717474235544$

4)

$\smallint \frac{(x+2) }{\sqrt{(x+3)}} dx$

Solution

$\smallint \frac{x+2}{\sqrt{x+3}} dx$

Substitute $u=x+3$

$\frac{du}{dx}=1 \to dx=du$

$=\smallint\frac{u-1}{\sqrt{u}}du$

Expand

$=\smallint[{\sqrt{u}}-\frac{1}{\sqrt{u}}]du$

Apply linearity

$=\smallint{\sqrt{u}}du - \smallint \frac{1}{\sqrt{u}}du$

Now solving

$=\smallint \sqrt{u} du$

Apply power rule:

$=\int u^ndu=\frac{u^{n+1}}{n+1}$ with $n=\frac{1}{2}$

$=\frac{2u^{\frac{3}{2}}}{3}$

Now solving

$\int\frac{1}{\sqrt{u}}du$

Apply power rule with $n=-\frac{1}{2}:$

$=2\sqrt{u}$

Plug in solved integrals

$=\smallint{\sqrt{u}}du - \smallint \frac{1}{\sqrt{u}}du$

$=\frac{2u^{\frac{3}{2}}}{3}-2\sqrt{u}$

Undo substitution $u=x+3:$

$=\frac{2(x+3)^{\frac{3}{2}}}{3}-2\sqrt{x+3}$

$=\frac{2(x+3)^{\frac{3}{2}}}{3}-2\sqrt{x+3} +C$

$=2[\frac{2(x+3)^{\frac{3}{2}}}{3}-\sqrt{x+3} ]+C$

$=\frac{2x(x+3)^{}}{3}+C$