Answer to Question #134547 in Calculus for sire

Question

1) ∫ x^4/x^10+100 dx
2) ∫ √x+9/x dx
3) ∫ dx/1+2e^x – e^-x
4) ∫ 2x^2dx/√9-x^2

Answer 1

1 $\int \frac{x^4}{x^{10}+100 } dx = \int \frac{x^4}{(x^{5})^2+10^2 } dx$

Let $x^5 = t \implies 5x^4dx = dt$

then

$\int \frac{x^4}{x^{10}+100 } dx = \frac{1}{5}\int \frac{dt}{t^{2}+10^2 } = \frac{1}{50}tan^{-1}(\frac{t}{10})$

$\int \frac{x^4}{x^{10}+100 } dx = \frac{1}{50}tan^{-1}(\frac{x^5}{10}) + C$

2 $\int \sqrt{x} + \frac{9}{x} dx = \frac{2}{3}x^{3/2} + 9 ln|x| + C$

3 $\int \frac{dx}{1+2e^{x} - e^{-x}} = \int \frac{e^{-x} dx}{e^{-x}+2-e^{-2x}}$

Let $e^{-x} = t \implies -e^{-x} dx = dt$

then $\int \frac{dx}{1+2e^{x} - e^{-x}} = \int \frac{ dt}{t+2-t^2} = - \int \frac{ dt}{t^2-t-2} = -\int \frac{ dt}{(t-\frac{1}{2})^2 - \frac{9}{4}}$

$= \frac{1}{3}ln|\frac{t+1}{t-2}| = \frac{1}{3}ln|\frac{e^{-x}-1}{e^{-x}-2}|$

4 $\int \frac{2x^2 dx}{\sqrt{9-x^2}} = -2\int \frac{9-x^2 -9 dx}{\sqrt{9-x^2}} = -2\int \sqrt{9-x^2} dx + 18\int \frac{1}{\sqrt{9-x^2}} dx$

$=-9sin^{-1}(\frac{x}{3}) - sin(2sin^{-1}(\frac{x}{3})) + 18sin^{-1}(\frac{x}{3}) + C$

Formula used:

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

$\int \frac{1}{x^2-a^2}dx=-\frac{1}{2a}\left(\ln \left|\frac{x}{a}+1\right|-\ln \left|\frac{x}{a}-1\right|\right)+C$

$\int \frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \left(\frac{x}{a}\right)+C$

$\int \sqrt{a^2-x^2}dx=\frac{1}{2}a^2\left(\arcsin \left(\frac{x}{a}\right)+\frac{1}{2}\sin \left(2\arcsin \left(\frac{x}{a}\right)\right)\right)+C$