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Question Number 81336 by mathmax by abdo last updated on 11/Feb/20

$$decomposeF\left(x\right)=\frac{1}{{(x-1)}^3{(x+3)}^7}$$$$anddetrmine\int F\left(x\right)dx$$$$2)calculate{\int }_2^{+\mathrm{\infty }}F\left(x\right)dx$$

Commented by Tony Lin last updated on 12/Feb/20

$$\int \frac{dx}{{(x-1)}^3{(x+3)}^7}$$$$letu=\frac{x+3}{x-1},\frac{du}{dx}=\frac{-4}{{(x-1)}^2}$$$$x=\frac{u+3}{u-1}$$$$\Rightarrow -\frac{1}{4}\int \frac{du}{\left(\frac{4}{u-1}\right){\left(\frac{4u}{u-1}\right)}^7}$$$$=-\frac{1}{4^9}\int \frac{{(u-1)}^8}{u^7}du$$$$=\int (u-8+\frac{28}{u}-\frac{56}{u^2}+\frac{70}{u^3}-\frac{56}{u^4}+\frac{28}{u^5}-\frac{8}{u^6}+\frac{1}{u^7})du$$$$\times (-\frac{1}{4^9})$$$$=(\frac{u^2}{2}-8u+28ln\mid u\mid +\frac{56}{u}-\frac{35}{u^2}+\frac{56}{3u^3}-\frac{7}{u^4}$$$$+\frac{8}{5u^5}-\frac{1}{6u^6})\times (-\frac{1}{262144})+c$$$$plugu=\frac{x+3}{x-1}in$$$$\Rightarrow \int \frac{dx}{{(x-1)}^3{(x+3)}^7}$$$$=-\frac{1}{262144}[\frac{{\left(\frac{x+3}{x-1}\right)}^2}{2}-\frac{8(x+3)}{x-1}+28ln\mid \frac{x+3}{x-1}\mid$$$$+\frac{56(x-1)}{x+3}-35{\left(\frac{x-1}{x+3}\right)}^2+\frac{56}{3}{\left(\frac{x-1}{x+3}\right)}^3-$$$$7{\left(\frac{x-1}{x+3}\right)}^4+\frac{8}{5}{\left(\frac{x-1}{x+3}\right)}^5-\frac{1}{6}{\left(\frac{x-1}{x+3}\right)}^6]+c$$

Commented by Tony Lin last updated on 12/Feb/20

$$\int \frac{dx}{{(x+a)}^m{(x+b)}^n}$$$$letu=\frac{x+b}{x+a},\frac{du}{dx}=\frac{a-b}{{(x+a)}^2}$$$$x=\frac{b-au}{u-1}$$$$\Rightarrow \frac{1}{a-b}\int \frac{du}{{\left(\frac{b-a}{u-1}\right)}^{m-2}{\left[\frac{(b-a)u}{u-1}\right]}^n}$$$$=-\frac{1}{{(b-a)}^{m+n-1}}\int \frac{{(u-1)}^{m+n-2}}{u^n}du$$$$=\int \frac{\underset{r=0}{\sum ^{m+n-2}}{C}_r^{m+n-2}{\left(\frac{x+b}{x+a}\right)}^{r-n}{(-1)}^{m+n-2-r}}{-{(b-a)}^{m+n-1}}d\left(\frac{x+b}{x+a}\right)$$

Commented by mathmax by abdo last updated on 12/Feb/20

$$thankyousir.$$

Commented by MJS last updated on 12/Feb/20

$$\mathrm{on}\mathrm{my}\mathrm{path}\mathrm{the}\mathrm{hard}\mathrm{work}\mathrm{is}\mathrm{solving}\mathrm{the}$$$$\mathrm{system}\mathrm{for}\mathrm{the}\mathrm{coefficients}{c}_j$$$$\mathrm{on}\mathrm{your}\mathrm{path}\mathrm{it}\mathrm{is}\mathrm{to}\mathrm{compose}\mathrm{all}\mathrm{those}\mathrm{fractions}$$$$\mathrm{if}\mathrm{needed}$$

Commented by mathmax by abdo last updated on 12/Feb/20

$$\int F\left(x\right)dx=\int \frac{dx}{{(x-1)}^3{(x+3)}^7}=\int \frac{dx}{{\left(\frac{x-1}{x+3}\right)}^3{(x+3)}^{10}}changement$$$$\frac{x-1}{x+3}=tgivex-1=tx+3t\Rightarrow (1-t)x=3t+1\Rightarrow x=\frac{3t+1}{1-t}\Rightarrow$$$$x+3=3+\frac{3t+1}{1-t}=\frac{3-3t+3t+1}{1-t}=\frac{4}{1-t}\Rightarrow dx=\frac{4}{{(1-t)}^2}\Rightarrow$$$$\int F\left(x\right)dx=\int \frac{4dt}{{(t-1)}^2{t}^3{\left(\frac{4}{1-t}\right)}^{10}}=4\int \frac{{(t-1)}^{10}}{4^{10}{(t-1)}^2{t}^3}dt$$$$=\frac{1}{4^9}\int \frac{{(t-1)}^8}{t^3}dt=\frac{1}{4^9}\int \frac{\sum _{k=0}^8{C}_8^k{t}^k{(-1)}^{8-k}}{t^3}dt$$$$=\frac{1}{4^9}\int \frac{1-C_8^{1}t+C_8^2{t}^2-C_8^3{t}^3+C_8^4{t}^4-C_8^5{t}^5+C_8^6{t}^6-C_8^7{t}^7+C_8^8{t}^8}{t^3}dt$$$$=\frac{1}{4^9}\int \frac{dt}{t^3}-\frac{C_8^1}{4^9}\int \frac{dt}{t^2}+\frac{C_8^2}{4^9}\int \frac{dt}{t}-\frac{C_8^3}{4^9}\int dt\frac{C_8^4}{4^9}\int tdt-\frac{C_8^5}{4^9}\int t^{2}dt$$$$+\frac{C_8^6}{4^9}\int t^{3}dt-\frac{C_8^7}{4^9}\int t^{4}dt+\frac{C_8^8}{4^9}\int t^{5}dt$$$$=-\frac{1}{2\times 4^9\times t^2}+\frac{8}{4^9\times t}+\frac{C_8^2}{4^9}ln\mid t\mid -\frac{C_8^3}{4^9}t+\frac{C_8^4}{2.4^9}{t}^2-\frac{C_8^5}{3.4^9}{t}^3+\frac{C_8^6}{4^{10}}{t}^4$$$$-\frac{C_8^7}{5.4^9}{t}^5+\frac{1}{6.4^9}{t}^6+C$$$$=-\frac{1}{2.4^9}{\left(\frac{x+3}{x-1}\right)}^2+\frac{8}{4^9}\left(\frac{x+3}{x-1}\right)+\frac{C_8^2}{4^9}ln\mid \frac{x-1}{x+3}\mid -\frac{C_8^3}{4^9}\left(\frac{x-1}{x+3}\right)$$$$+\frac{C_8^4}{2.4^9}{\left(\frac{x-1}{x+3}\right)}^2-\frac{C_8^5}{3.4^9}{\left(\frac{x-1}{x+3}\right)}^3+\frac{C_8^6}{4^{10}}{\left(\frac{x-1}{x+3}\right)}^4-\frac{C_8^7}{5.4^9}{\left(\frac{x-1}{x+3}\right)}^5+\frac{1}{6.4^9}{\left(\frac{x-1}{x+3}\right)}^6+C$$$$$$

Commented by mathmax by abdo last updated on 12/Feb/20

$${\int }_2^{+\mathrm{\infty }}F\left(x\right)dx=-\frac{1}{2.4^9}{\left[{\left(\frac{x+3}{x-1}\right)}^2\right]}_2^{+\mathrm{\infty }}+\frac{8}{4^9}{\left[\left(\frac{x+3}{x-1}\right)\right]}_2^{+\mathrm{\infty }}+\frac{C_8^2}{4^9}{[ln\mid \frac{x-1}{x+3}\mid ]}_2^{+\mathrm{\infty }}$$$$-\frac{C_8^3}{4^9}{\left[\left(\frac{x-1}{x+3}\right)\right]}_2^{+\mathrm{\infty }}+\frac{C_8^4}{2.4^9}{\left[{\left(\frac{x-1}{x+3}\right)}^2\right]}_2^{+\mathrm{\infty }}-\frac{C_8^5}{3.4^9}{\left[{\left(\frac{x-1}{x+3}\right)}^3\right]}_2^{+\mathrm{\infty }}$$$$+\frac{C_8^6}{4^{10}}{\left[{\left(\frac{x-1}{x+3}\right)}^4\right]}_2^{+\mathrm{\infty }}-\frac{C_8^7}{5.4^9}{\left[{\left(\frac{x-1}{x+3}\right)}^5\right]}_2^{+\mathrm{\infty }}+\frac{1}{6.4^9}{\left[{\left(\frac{x-1}{x+3}\right)}^6\right]}_2^{+\mathrm{\infty }}$$$$=-\frac{1}{2.4^9}\{1-5^2\}+\frac{8}{4^9}\{1-5\}+\frac{C_8^2}{4^9}\{-ln\left(\frac{1}{5}\right)\}$$$$-\frac{C_8^3}{4^9}\{1-\frac{1}{5}\}+\frac{C_8^4}{2.4^9}\{1-{\left(\frac{1}{5}\right)}^2\}-\frac{C_8^5}{3.4^9}\{1-{\left(\frac{1}{5}\right)}^3\}$$$$+\frac{C_8^6}{4^{10}}\{1-{\left(\frac{1}{5}\right)}^4\}-\frac{C_8^7}{5.4^9}\{1-{\left(\frac{1}{5}\right)}^5\}+\frac{1}{6.4^9}\{1-{\left(\frac{1}{5}\right)}^6\}$$

Commented by mathmax by abdo last updated on 12/Feb/20

$$theavantageofthismethodisfindingintegralwithoutpassing$$$$bydecomposition..!$$

Answered by MJS last updated on 12/Feb/20

$${\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\mathrm{again}({\mathrm{it}}^{\prime }\mathrm{s}\mathrm{faster}\mathrm{in}\mathrm{my}$$$$\mathrm{opinion})$$$$\int \frac{dx}{{(x+3)}^7{(x-1)}^3}$$$$\int \frac{P\left(x\right)}{Q\left(x\right)}dx=\frac{P_1\left(x\right)}{Q_1\left(x\right)}+\int \frac{P_2\left(x\right)}{Q_2\left(x\right)}dx$$$$Q_1\left(x\right)=\gcd (Q\left(x\right),Q^{{}^{\prime }}\left(x\right))={(x+3)}^6{(x-1)}^{2}oo$$$$Q_2\left(x\right)=\frac{Q\left(x\right)}{Q_1\left(x\right)}=(x+3)(x-1)$$$$\frac{P\left(x\right)}{Q\left(x\right)}=\frac{d}{dx}\left[\frac{P_1\left(x\right)}{Q_1\left(x\right)}\right]+\frac{P_2\left(x\right)}{Q_2\left(x\right)}$$$$\frac{1}{{(x+3)}^7{(x-1)}^3}=\frac{d}{dx}\left[\frac{\underset{j=0}{\sum ^7}{c}_j{x}^j}{{(x+3)}^6{(x-1)}^2}\right]+\frac{d_{1}x+d_0}{(x+3)(x-1)}$$$$\Rightarrow$$$$P_1\left(x\right)=\frac{105x^7+1575x^6+9065x^5+23135x^4+15491x^3-38339x^2-56405x+14653}{245760}$$$$P_2\left(x\right)=\frac{7}{16384}$$$$\Rightarrow \mathrm{we}\mathrm{only}\mathrm{have}\mathrm{to}\mathrm{solve}\frac{7}{16384}\int \frac{dx}{(x+3)(x-1)}=$$$$=\frac{7}{65536}\ln \mid \frac{x-1}{x+3}\mid$$$$\int \frac{dx}{{(x+3)}^7{(x-1)}^3}=$$$$=\frac{105x^7+1575x^6+9065x^5+23135x^4+15491x^3-38339x^2-56405x+14653}{245760{(x+3)}^6{(x-1)}^2}+\frac{7}{65536}\ln \mid \frac{x-1}{x+3}\mid +C$$$$\underset{2}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{{(x+3)}^7{(x-1)}^3}=\frac{-517516+328125\ln 5}{3072000000}$$

Commented by john santu last updated on 12/Feb/20

$${ostrogradski}^{\prime }smethodcanuseany$$$$typeintegral?$$

Commented by msup trace by abdo last updated on 12/Feb/20

$$thankyousir.$$

Commented by MJS last updated on 12/Feb/20

$$\mathrm{no}.P\left(x\right)\mathrm{and}Q\left(x\right)\mathrm{must}\mathrm{be}\mathrm{polynomes}.$$$$\mathrm{the}\mathrm{degree}\mathrm{of}P\left(x\right)\mathrm{must}\mathrm{be}\mathrm{smaller}\mathrm{than}$$$$\mathrm{the}\mathrm{degree}\mathrm{of}Q\left(x\right)\mathrm{and}\mathrm{all}\mathrm{coefficients}\mathrm{must}$$$$\mathrm{be}\mathrm{real}$$$$\mathrm{the}\mathrm{degree}\mathrm{of}{P}_1\left(x\right)\mathrm{will}\mathrm{then}\mathrm{be}\mathrm{smaller}\mathrm{than}$$$$\mathrm{the}\mathrm{degree}\mathrm{of}{Q}_1\left(x\right),\mathrm{same}\mathrm{for}{P}_2\left(x\right)\mathrm{and}{Q}_2\left(x\right)$$$$\mathrm{I}\mathrm{think}\mathrm{you}\mathrm{can}\mathrm{find}\mathrm{information}\mathrm{on}\mathrm{wikipedia}$$

Commented by jagoll last updated on 12/Feb/20

$$\mathrm{thank}\mathrm{you}\mathrm{mister}$$

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