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Question Number 50158 by cesar.marval.larez@gmail.com last updated on 14/Dec/18

Commented by cesar.marval.larez@gmail.com last updated on 14/Dec/18

$$iampracticingveryhard$$

Answered by afachri last updated on 14/Dec/18

$$\left(4\right)\int \frac{2x^2+8x+6}{x-3}dx=....$$$$\mathbf{let}x-3=u\Rightarrow dx=du$$$$x=u+3\mathbf{then}\mathbf{subtitute}\mathbf{to}\mathbf{eq};$$$$\int \frac{2{(u+3)}^2+8(u+3)+6}{u}du=\int \frac{2u^2+20u+48}{u}$$$$=\int 2u+20+\frac{48}{u}du$$$$=u^2+20u+48\ln u+C$$$$={(x-3)}^2+20(x-3)+48\ln (x-3)+C$$$$=x^2+14x-51+48\ln (x-3)+C$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

$$4)\int \frac{2x^2+3x+6}{x-3}dx$$$$t=x-3$$$$\int \frac{2{(t+3)}^2+3(t+3)+6}{t}dt$$$$\int \frac{2t^2+12t+18+3t+9+6}{t}dt$$$$\int \frac{2t^2+15t+33}{t}dt$$$$2\int tdt+15\int dt+33\int \frac{dt}{t}$$$$t^2+15t+33lnt+c$$$${(x-3)}^2+15(x-3)+33ln(x-3)+c$$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

$$5)\int {cos}^{4}6x(-6sin6x)dx$$$$t=cos6xdt=-6sin6xdx$$$$\int t^{4}dt$$$$\frac{t^5}{5}+c$$$$\frac{{\left(cos6x\right)}^5}{5}+c$$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

$$6)\int {sin}^{3}xdx$$$$\int (1-{cos}^{2}x)sinxdx$$$$t=cosxdt=-sinxdx$$$$\int (1-t^2)\times -dt$$$$=(-1)(t-\frac{t^3}{3})+c$$$$=\frac{t^3}{3}-t+c$$$$=\frac{{\left(cosx\right)}^3}{3}-cosx+c$$

Commented by cesar.marval.larez@gmail.com last updated on 14/Dec/18

$$myfriendis{\mathit{s}\mathit{e}\mathit{n}}^3\mathit{a}\mathit{x}$$

Commented by afachri last updated on 14/Dec/18

$$\mathrm{you}\mathrm{can}\mathrm{solve}\mathrm{it}\mathrm{over}\mathrm{and}\mathrm{do}\mathrm{like}$$$$\mathrm{Mr}\mathrm{Chaudri}\mathrm{did}.\mathrm{and}\mathrm{the}\mathrm{result}$$$$\mathrm{is}:$$$$\left(\frac{\cos {}^3\left(ax\right)-3\cos \left(ax\right)}{3a}\right)+C\mathrm{not}\mathrm{too}\mathrm{far}$$$${\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{it}??$$$${\mathrm{don}}^{\prime }\mathrm{t}\mathrm{be}\mathrm{doubt}\mathrm{to}\mathrm{try}\mathrm{my}\mathrm{friend}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

$$\int {cos}^2\frac{x}{2}dx$$$$\int \frac{1+cosx}{2}dx$$$$\frac{1}{2}\int (1+cosx)dx$$$$=\frac{1}{2}(x+sinx)+c$$

Commented by cesar.marval.larez@gmail.com last updated on 14/Dec/18

$$Iknewthisbutiwasnotsecure$$$$thankusirhaha$$

Answered by afachri last updated on 15/Dec/18

$$\left(9\right)\int 5x.e^{\frac{4x}{5}}dx=...$$$$\mathrm{Can}\mathrm{be}\mathrm{solved}\mathrm{with}\mathrm{partial}\mathrm{integral};$$$$\int udv=uv-\int vdu$$$$$$$$\mathbf{let}\mathit{u}=5x\Rightarrow du=5dx$$$$dv=e^{\frac{4x}{5}}dx\Rightarrow v=\frac{5}{4}{e}^{\frac{4x}{5}}$$$$\mathbf{so}uv-\int vdu=(5x.\frac{5}{4}{e}^{\frac{4x}{5}})-\int \frac{5}{4}{e}^{\frac{4x}{5}}\left(5dx\right)$$$$=\frac{25x}{4}{e}^{\frac{4x}{5}}-(\frac{5\times 5}{4\times 4}e{}^{\frac{4x}{5}}.5)+C$$$$=\frac{25e^{\frac{4x}{5}}}{4}(x-\frac{5}{4})+C$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

$$7)\int 2x^3{e}^{-\frac{x}{3}}dx$$$$2x^3\int e^{-\frac{x}{3}}dx-\int [\frac{d}{dx}\left(2x^3\right)\int e^{-\frac{x}{3}}dx]dx$$$$=2x^3\times \frac{e^{\frac{-x}{3}}}{\frac{-1}{3}}-\int 6x^2\times \frac{e^{-\frac{x}{3}}}{\frac{-1}{3}}dx$$$$=-6x^3{e}^{\frac{-x}{3}}+18I_2$$$$I_2=\int x^2{e}^{\frac{-x}{3}}dx$$$$=x^2\times \frac{e^{\frac{-x}{3}}}{\frac{-1}{3}}-\int [\frac{{dx}^2}{dx}\int e^{\frac{-x}{3}}dx]dx$$$$=-3x^2{e}^{\frac{-x}{3}}-\int 2x\times \frac{e^{\frac{-x}{3}}}{\frac{-1}{3}}dx$$$$=-3x^2{e}^{\frac{-x}{3}}+6\int {xe}^{\frac{-x}{3}}dx$$$$=-3x^2{e}^{\frac{-x}{3}}+6I_3$$$$I_3=\int {xe}^{\frac{-x}{3}}dx$$$$=x\times \frac{e^{-\frac{x}{3}}}{\frac{-1}{3}}-\int [\frac{dx}{dx}\int e^{\frac{-x}{3}}dx]dx$$$$=\frac{-3{xe}^{\frac{-x}{3}}}{1}-\int \frac{e^{\frac{-x}{3}}}{\frac{-1}{3}}dx$$$$=\frac{-3{xe}^{\frac{-x}{3}}}{1}+3\times \frac{e^{\frac{-x}{3}}}{\frac{-1}{3}}+C$$$$=-3{xe}^{\frac{-x}{3}}-9e^{-\frac{x}{3}}+C$$$$plsdorestbyputting{I}_{2}and{I}_3$$

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