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2014-12-18T20:23:53+05:30

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The integral can be found step by step as a series as :

I= \int\limits^{}_{} {e^{x^3}*1} \, dx,\ \ \ u=e^{x^3},\ v'=1, \ v=x\\\\I=e^{x^3}*x- \int\limits^{}_{} ({e^{x^3}*3x^2})(x) \, dx\\\\I=x*e^{x^3}-3 \int\limits^{}_{} {e^{x^3}*x^3} \, dx\\ v'=x^3,\ v=x^4/4,\ u=e^{x^3},\ u'=3x^2*e^{x^3}\\\\ I=x*e^{x^3}-3 [ e^{x^3}*x^4/4 -  \int\limits^{}_{} {3x^2e^{x^3}x^4/4} \, dx ]\\

I=x*e^{x^3}-\frac{3}{4}x^4e^{x^3}+\frac{9}{4} \int\limits^{}_{} {x^6e^{x^3}} \, dx \\\\ \int\limits^{}_{} {x^6e^{x^3}} \, dx =e^{x^3}*x^7/7- \int\limits^{}_{} {3x^2e^{x^3}x^7/7} \, dx=e^{x^3}*x^7/7- \int\limits^{}_{} {3x^9e^{x^3}/7} \, dx\\\\\int\limits^{}_{} {x^9e^{x^3}} \, dx=e^{x^3}*x^{10}/10- \int\limits^{}_{} {3x^{12}e^{x^3}/10} \, dx\\\\\int\limits^{}_{} {x^{12}e^{x^3}} \, dx=e^{x^3}x^{13}/13- \int\limits^{}_{} {3x^{15}e^{x^3}/13} \, dx \\


I=e^{x^3}[x-\frac{3}{4}x^4+\frac{9}{28}x^7-\frac{27}{280}x^{10}+\frac{81}{280*13}x^{13}]-\frac{243}{280*13} \int\limits^{}_{} {x^{15}e^{x^3}} \, dx\\


We can find the expansion like the Taylor series for integral of  e^x^3 wrt x.


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