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Question Number 104928 by bobhans last updated on 24/Jul/20

∫ (e^x −(2x+3)^4 )^3  dx

$$\int\:\left({e}^{{x}} −\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{4}} \right)^{\mathrm{3}} \:{dx} \\ $$

Commented by kaivan.ahmadi last updated on 24/Jul/20

∫(e^(3x) −3e^(2x) (2x+3)^4 +3e^x (2x+3)^8 −(2x+3)^(12) )dx=  (1/3)e^(3x) −(1/(26))(2x+3)^(13) −3∫e^(2x) (2x+3)^4 dx+3∫e^x (2x+3)^8 dx=  we must find two integrals.we find one of  them and similarly you can find another.

$$\int\left({e}^{\mathrm{3}{x}} −\mathrm{3}{e}^{\mathrm{2}{x}} \left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{4}} +\mathrm{3}{e}^{{x}} \left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{8}} −\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{12}} \right){dx}= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}{e}^{\mathrm{3}{x}} −\frac{\mathrm{1}}{\mathrm{26}}\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{13}} −\mathrm{3}\int{e}^{\mathrm{2}{x}} \left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{4}} {dx}+\mathrm{3}\int{e}^{{x}} \left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{8}} {dx}= \\ $$$${we}\:{must}\:{find}\:{two}\:{integrals}.{we}\:{find}\:{one}\:{of} \\ $$$${them}\:{and}\:{similarly}\:{you}\:{can}\:{find}\:{another}. \\ $$$$ \\ $$$$ \\ $$

Commented by kaivan.ahmadi last updated on 24/Jul/20

Commented by kaivan.ahmadi last updated on 24/Jul/20

∫e^(2x) (2x+3)^4 dx=8e^(2x) (2x+3)^3 −24e^(2x) (2x+3)^(2+)   48e^(2x) (2x+3)−48e^(2x)

$$\int{e}^{\mathrm{2}{x}} \left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{4}} {dx}=\mathrm{8}{e}^{\mathrm{2}{x}} \left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{3}} −\mathrm{24}{e}^{\mathrm{2}{x}} \left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}+} \\ $$$$\mathrm{48}{e}^{\mathrm{2}{x}} \left(\mathrm{2}{x}+\mathrm{3}\right)−\mathrm{48}{e}^{\mathrm{2}{x}} \\ $$

Commented by malwaan last updated on 25/Jul/20

sir ahmadi  the first term must be  (1/2) e^(2x)  (2x + 3 )^4  ...  and so on ....

$$\mathrm{sir}\:{ahmadi} \\ $$$${the}\:{first}\:{term}\:{must}\:{be} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}} \:\left(\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{3}\:\right)^{\mathrm{4}} \:... \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{so}}\:\boldsymbol{{on}}\:.... \\ $$

Commented by malwaan last updated on 25/Jul/20

∫e^x (2x+3)^8 dx = e^x {(2x+3)^8   −16(2x+3)^7 +224(2x+3)^6   −2688(2x+3)^5   +26880(2x+3)^4   −215040(2x+3)^3   +860160(2x+3)^2   −1720320(2x+3)  +3440640}

$$\int\boldsymbol{{e}}^{\boldsymbol{{x}}} \left(\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\right)^{\mathrm{8}} \boldsymbol{{dx}}\:=\:\boldsymbol{{e}}^{\boldsymbol{{x}}} \left\{\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\right)^{\mathrm{8}} \right. \\ $$$$−\mathrm{16}\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\right)^{\mathrm{7}} +\mathrm{224}\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\right)^{\mathrm{6}} \\ $$$$−\mathrm{2688}\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\right)^{\mathrm{5}} \\ $$$$+\mathrm{26880}\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\right)^{\mathrm{4}} \\ $$$$−\mathrm{215040}\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\right)^{\mathrm{3}} \\ $$$$+\mathrm{860160}\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\right)^{\mathrm{2}} \\ $$$$−\mathrm{1720320}\left(\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\right) \\ $$$$\left.+\mathrm{3440640}\right\} \\ $$

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