Exercise 7.4 Detailed Solutions

Let $x^3 = t \implies 3x^2 dx = dt$.
Integral becomes $\int \frac{dt}{t^2+1}$.
Using $\int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}(\frac{x}{a})$:
$= \tan^{-1} t + C = \tan^{-1}(x^3) + C$.

Write as $\int \frac{1}{\sqrt{1+(2x)^2}} dx$. Let $2x = t \implies dx = dt/2$.
$\frac{1}{2} \int \frac{dt}{\sqrt{1+t^2}} = \frac{1}{2}\log|t+\sqrt{1+t^2}| + C$.
$= \frac{1}{2}\log|2x+\sqrt{1+4x^2}| + C$.

Let $2-x = t \implies -dx = dt$.
$-\int \frac{dt}{\sqrt{t^2+1}} = -\log|t+\sqrt{t^2+1}| + C$.
$= \log|\frac{1}{2-x+\sqrt{x^2-4x+5}}| + C$.

$= \int \frac{1}{\sqrt{3^2-(5x)^2}} dx$. Let $5x=t \implies dx=dt/5$.
$\frac{1}{5} \int \frac{dt}{\sqrt{3^2-t^2}} = \frac{1}{5}\sin^{-1}(\frac{t}{3}) + C$.
$= \frac{1}{5}\sin^{-1}(\frac{5x}{3}) + C$.

$= \int \frac{3x}{1+(\sqrt{2}x^2)^2} dx$. Let $\sqrt{2}x^2 = t \implies 2\sqrt{2}x dx = dt$.
$x dx = \frac{dt}{2\sqrt{2}}$.
$\frac{3}{2\sqrt{2}} \int \frac{dt}{1+t^2} = \frac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2}x^2) + C$.

$= \int \frac{x^2}{1-(x^3)^2} dx$. Let $x^3=t \implies 3x^2 dx = dt$.
$\frac{1}{3} \int \frac{dt}{1-t^2} = \frac{1}{3} \cdot \frac{1}{2}\log|\frac{1+t}{1-t}| + C$.
$= \frac{1}{6}\log|\frac{1+x^3}{1-x^3}| + C$.

Split integrals: $\int \frac{x}{\sqrt{x^2-1}} dx – \int \frac{1}{\sqrt{x^2-1}} dx$.
$I_1$: Let $x^2-1=t \implies 2x dx = dt$. $\frac{1}{2}\int t^{-1/2} dt = \sqrt{x^2-1}$.
$I_2$: Standard formula $\log|x+\sqrt{x^2-1}|$.
Result: $\sqrt{x^2-1} – \log|x+\sqrt{x^2-1}| + C$.

Let $x^3=t \implies 3x^2 dx = dt$.
$\frac{1}{3} \int \frac{dt}{\sqrt{t^2+(a^3)^2}} = \frac{1}{3}\log|t+\sqrt{t^2+a^6}| + C$.
$= \frac{1}{3}\log|x^3+\sqrt{x^6+a^6}| + C$.

Let $\tan x = t \implies \sec^2 x dx = dt$.
$\int \frac{dt}{\sqrt{t^2+2^2}} = \log|t+\sqrt{t^2+4}| + C$.
$= \log|\tan x + \sqrt{\tan^2 x + 4}| + C$.

Completing Square: $x^2+2x+2 = (x+1)^2+1$.
$\int \frac{dx}{\sqrt{(x+1)^2+1}} = \log|(x+1)+\sqrt{x^2+2x+2}| + C$.

$= \frac{1}{9} \int \frac{dx}{x^2+\frac{2}{3}x+\frac{5}{9}}$.
$= \frac{1}{9} \int \frac{dx}{(x+\frac{1}{3})^2 + (\frac{2}{3})^2}$.
$= \frac{1}{9} \cdot \frac{1}{2/3} \tan^{-1}(\frac{x+1/3}{2/3}) = \frac{1}{6}\tan^{-1}(\frac{3x+1}{2}) + C$.

$7-(x^2+6x) = 7-(x^2+6x+9)+9 = 16-(x+3)^2$.
$\int \frac{dx}{\sqrt{4^2-(x+3)^2}} = \sin^{-1}(\frac{x+3}{4}) + C$.

$\sqrt{x^2-3x+2} = \sqrt{(x-\frac{3}{2})^2 – (\frac{1}{2})^2}$.
$\int \frac{dx}{\sqrt{X^2-a^2}} = \log|x-\frac{3}{2} + \sqrt{x^2-3x+2}| + C$.

$8-(x^2-3x) = 8-(x^2-3x+\frac{9}{4})+\frac{9}{4} = \frac{41}{4} – (x-\frac{3}{2})^2$.
$\int \frac{dx}{\sqrt{(\frac{\sqrt{41}}{2})^2 – (x-\frac{3}{2})^2}} = \sin^{-1}(\frac{2x-3}{\sqrt{41}}) + C$.

Denominator is $\sqrt{x^2-(a+b)x+ab}$. Standard form $\log|x + \sqrt{x^2-a^2}|$.
Result: $\log|x-\frac{a+b}{2} + \sqrt{(x-a)(x-b)}| + C$.

Derivative of $2x^2+x-3$ is $4x+1$.
Let $2x^2+x-3=t \implies (4x+1)dx = dt$.
$\int t^{-1/2} dt = 2\sqrt{t} = 2\sqrt{2x^2+x-3} + C$.

Split: $\int \frac{x}{\sqrt{x^2-1}} dx + 2\int \frac{1}{\sqrt{x^2-1}} dx$.
$I_1$: Let $x^2-1=t \implies \sqrt{x^2-1}$.
$I_2$: $2\log|x+\sqrt{x^2-1}|$.
Result: $\sqrt{x^2-1} + 2\log|x+\sqrt{x^2-1}| + C$.

Let $5x-2 = A\frac{d}{dx}(1+2x+3x^2) + B = A(6x+2) + B$.
$6A = 5 \implies A=5/6$. $2A+B=-2 \implies B=-11/3$.
$I_1 = \frac{5}{6} \int \frac{6x+2}{3x^2+2x+1} dx = \frac{5}{6}\log|3x^2+2x+1|$.
$I_2 = -\frac{11}{3} \int \frac{dx}{3(x^2+\frac{2}{3}x+\frac{1}{3})} = -\frac{11}{9} \int \frac{dx}{(x+\frac{1}{3})^2 + (\frac{\sqrt{2}}{3})^2}$.
$I_2 = -\frac{11}{9} \cdot \frac{3}{\sqrt{2}} \tan^{-1}(\frac{3x+1}{\sqrt{2}})$.
Ans: $\frac{5}{6}\log|3x^2+2x+1| – \frac{11}{3\sqrt{2}}\tan^{-1}(\frac{3x+1}{\sqrt{2}}) + C$.

Denominator: $\sqrt{x^2-9x+20}$. Derivative is $2x-9$.
$6x+7 = A(2x-9) + B \implies 2A=6 \to A=3; -9A+B=7 \to B=34$.
$I_1 = 3 \int \frac{2x-9}{\sqrt{x^2-9x+20}} dx = 6\sqrt{x^2-9x+20}$.
$I_2 = 34 \int \frac{dx}{\sqrt{(x-4.5)^2 – (0.5)^2}} = 34\log|x-4.5+\sqrt{x^2-9x+20}|$.

Derivative of $4x-x^2$ is $4-2x$.
$x+2 = -\frac{1}{2}(4-2x) + 4$.
$I_1 = -\frac{1}{2} \cdot 2\sqrt{4x-x^2} = -\sqrt{4x-x^2}$.
$I_2 = 4 \int \frac{dx}{\sqrt{2^2-(x-2)^2}} = 4\sin^{-1}(\frac{x-2}{2})$.
Ans: $-\sqrt{4x-x^2} + 4\sin^{-1}(\frac{x-2}{2}) + C$.

Derivative is $2x+2$.
$x+2 = \frac{1}{2}(2x+2) + 1$.
$I_1 = \frac{1}{2} \cdot 2\sqrt{x^2+2x+3} = \sqrt{x^2+2x+3}$.
$I_2 = \int \frac{dx}{\sqrt{(x+1)^2+2}} = \log|x+1+\sqrt{x^2+2x+3}|$.
Ans: $\sqrt{x^2+2x+3} + \log|x+1+\sqrt{x^2+2x+3}| + C$.

Derivative is $2x-2$.
$x+3 = \frac{1}{2}(2x-2) + 4$.
$I_1 = \frac{1}{2}\log|x^2-2x-5|$.
$I_2 = 4 \int \frac{dx}{(x-1)^2-(\sqrt{6})^2} = \frac{4}{2\sqrt{6}}\log|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|$.
Ans: $\frac{1}{2}\log|x^2-2x-5| + \frac{2}{\sqrt{6}}\log|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}| + C$.

Derivative is $2x+4$.
$5x+3 = \frac{5}{2}(2x+4) – 7$.
$I_1 = \frac{5}{2} \cdot 2\sqrt{x^2+4x+10} = 5\sqrt{x^2+4x+10}$.
$I_2 = -7 \int \frac{dx}{\sqrt{(x+2)^2+6}} = -7\log|x+2+\sqrt{x^2+4x+10}|$.

$= \int \frac{dx}{(x+1)^2+1} = \tan^{-1}(x+1) + C$.

Factor 4 out of sqrt: $\frac{1}{2} \int \frac{dx}{\sqrt{\frac{9}{4}x-x^2}}$.
Complete square inside: $-[x^2-\frac{9}{4}x] = (\frac{9}{8})^2 – (x-\frac{9}{8})^2$.
$\frac{1}{2} \sin^{-1}(\frac{x-9/8}{9/8}) = \frac{1}{2}\sin^{-1}(\frac{8x-9}{9}) + C$.