Integration Problems

[rjs123]

1. ?(?4x² + 9)/x4
using u = atan? x = 3/2tan? dx = 3/2sec²?
I plugged x in and am stuck on simplifying

2. ?x²/?25 - x²)
u = asin? x = 5sin? dx = 5cos?
Again, plugged x into original equation cannot simplify

3. ?2x^3 - 4x - 8/(x² - x)(x² + 4)
got stuck midway through this problem

2x^3 - 4x - 8 = A/x + B/x+1 + Cx +D/(x² + 4)

A = 2
B = -2

 

[galactus]

2. ?x²/?25 - x²)
u = asin? x = 5sin? dx = 5cos?
Again, plugged x into original equation cannot simplify

Your sub is correct. Assuming you have to use trig sub.

\(\displaystyle \int\frac{25sin^{2}\theta}{\sqrt{25(1-sin^{2}\theta)}}5cos\theta d\theta\)

\(\displaystyle \int\frac{25sin^{2}\theta}{\sqrt{25cos^{2}\theta}}5cos\theta d\theta\)

\(\displaystyle 25\int sin^{2}\theta d\theta=25\int\frac{1-cos2\theta}{2}d\theta\)

Now, integrate and resub. Remember, after integrating, you resub \(\displaystyle \theta=sin^{-1}(\frac{x}{5})\)

3. ?2x^3 - 4x - 8/(x² - x)(x² + 4)
got stuck midway through this problem

2x^3 - 4x - 8 = A/x + B/x+1 + Cx +D/(x² + 4)

A = 2
B = -2

Expanding with partial fractions, one gets:

\(\displaystyle \int\frac{2x}{x^{2}+4}dx+\int\frac{4}{x^{2}+4}dx-\int\frac{2}{x-1}dx+\int\frac{2}{x}dx\)

Now, integrate each term.

 

[soroban]

Hello, rjs123"!

It would help if you showed us your work.
I don't know what "I plugged in x" means.

\(\displaystyle \displaystyle 1.\;\int \frac{\sqrt{4x^2+9}}{x^4}\,dx\)

\(\displaystyle \text{Let: }\:2x \:=\:3\tan\theta \quad\Rightarrow\quad x \:=\:\tfrac{3}{2}\tan\theta \quad\Rightarrow\quad dx \:=\:\tfrac{3}{2}\sec^2\!\theta\,d\theta\)


\(\displaystyle \displaystyle \text{Substitute: }\:\int \frac{3\sec\theta}{(\frac{3}{2}\tan\theta)^4}\,(\tfrac{3}{2}\sec^2\!\theta\,d\theta) \;\;=\;\;\tfrac{8}{9}\int\frac{\sec^3\!\theta}{\tan^4\!\theta}\,d\theta \;\;=\;\;\tfrac{8}{9}\int\frac{\frac{1}{\cos^3\!\theta}}{\frac{\sin^4\theta}{\cos^4\theta}}\,d\theta\)

. . . . . . \(\displaystyle \displaystyle =\;\;\tfrac{8}{9}\int\frac{\cos\theta\,d\theta}{\sin^4\theta} \;\;=\;\;\tfrac{8}{9}\int (\sin\theta)^{\text{-}4}(\cos\theta\,d\theta)\)


\(\displaystyle \text{Let: }\:u \,=\,\sin\theta \quad\Rightarrow\quad du \,=\,\cos\theta\,d\theta\)

\(\displaystyle \text{Substitute: }\: \tfrac{8}{9}\int u^{\text{-}4}du \;=\;-\tfrac{8}{27}u^{\text{-}3} + C\)

\(\displaystyle \text{Back-substitute: }\:-\tfrac{8}{27}(\sin\theta)^{\text{-}3} + C \;\;=\;\;-\tfrac{8}{27}\csc^3\theta + C\)


\(\displaystyle \text{Back-substitute: }\:\tan\theta \:=\:\frac{2x}{3} \:=\:\frac{opp}{adj}\)
. . \(\displaystyle \text{Then: }\:hyp \:=\:\sqrt{4x^2+9} \quad\Rightarrow\quad \csc\theta \:=\:\frac{hyp}{opp} \:=\:\frac{\sqrt{4x^2+9}}{2x}\)

\(\displaystyle \text{And we have: }\:-\frac{8}{27}\left(\frac{\sqrt{4x^2+9}}{2x}\right)^3 + C \;=\;-\frac{8}{27}\cdot\frac{(4x^2+9)^{\frac{3}{2}}}{8x^3} + C\)


. . . . \(\displaystyle \text{Answer: }\;-\frac{(4x^2+9)^{\frac{3}{2}}}{27x^3} + C\)