From Förberedande kurs i matematik 2

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-	{{NAVCONTENT_START}}	+	We have a product of two factors in the integrand, so an integration by parts does not seem unreasonable. There is nevertheless a problem as regards which factor should be differentiated and which should be integrated. If we choose to differentiate <math>x^3</math> (so as to reduce its exponent by 1), we need to find a primitive function for <math>e^{x^2}</math>, and how do we do that? If, on the other hand, we integrate <math>x^3</math> and differentiate <math>e^{x^2}</math>, we get
-	<center> [[Image:2_3_2b-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
-	{{NAVCONTENT_START}}	+	\int x^3\cdot e^{x^2}\,dx
-	<center> [[Image:2_3_2b-2(2).gif]] </center>	+	&= \frac{x^4}{4}\cdot e^{x^2} - \int\frac{x^4}{4}\cdot e^{x^2}2x\,dx\\[5pt]
-	{{NAVCONTENT_STOP}}	+	&= \frac{1}{4}x^{4}e^{x^2} - \frac{1}{2}\int x^5e^{x^2}\,dx
		+	\end{align}</math>}}
		+
		+	which just seems to make the integral harder. The solution is instead to substitute
		+	<math>u=x^2</math>. If we write the integral as
		+
		+	{{Displayed math||<math>\int\limits_0^1 x^3e^{x^2}\,dx = \int\limits_0^1 x^2e^{x^2}x\,dx</math>}}
		+
		+	we see that the expression "<math>x\,dx</math>" can be replaced by <math>du</math>
		+	and the rest of the integrand contains only <math>x</math> in the form of <math>x^2</math>. The substitution gives
		+
		+	{{Displayed math||<math>\begin{align}
		+	\int\limits_0^1 x^3e^{x^2}\,dx
		+	&= \int\limits_0^1 x^2e^{x^2}x\,dx\\[5pt]
		+	&= \left\{\begin{align}
		+	u &= x^2\\[5pt]
		+	du &= \bigl(x^2\bigr)'\,dx = 2x\,dx
		+	\end{align}\right\}\\[5pt]
		+	&= \int\limits_0^1 ue^u\tfrac{1}{2}\,du\\[5pt]
		+	&= \frac{1}{2}\int\limits_0^1 ue^u\,du\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	We can then calculate this integral by integration by parts, where we differentiate away the factor <math>u</math>,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\frac{1}{2}\int\limits_0^1 ue^u\,du
		+	&= \frac{1}{2}\Bigl[\ ue^u\ \Bigr]_0^1 - \frac{1}{2}\int\limits_0^1 1\cdot e^u\,du\\[5pt]
		+	&= \frac{1}{2}\bigl(1\cdot e^1-0\bigr) - \frac{1}{2}\Bigl[\ e^u\ \Bigr]_0^1\\[5pt]
		+	&= \frac{1}{2}e - \frac{1}{2}\bigl(e^1-e^0\bigr)\\[5pt]
		+	&= \frac{1}{2}e - \frac{1}{2}e + \frac{1}{2}\\[5pt]
		+	&= \frac{1}{2}\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

We have a product of two factors in the integrand, so an integration by parts does not seem unreasonable. There is nevertheless a problem as regards which factor should be differentiated and which should be integrated. If we choose to differentiate \displaystyle x^3 (so as to reduce its exponent by 1), we need to find a primitive function for \displaystyle e^{x^2}, and how do we do that? If, on the other hand, we integrate \displaystyle x^3 and differentiate \displaystyle e^{x^2}, we get

\displaystyle \begin{align} \int x^3\cdot e^{x^2}\,dx &= \frac{x^4}{4}\cdot e^{x^2} - \int\frac{x^4}{4}\cdot e^{x^2}2x\,dx\\[5pt] &= \frac{1}{4}x^{4}e^{x^2} - \frac{1}{2}\int x^5e^{x^2}\,dx \end{align}

which just seems to make the integral harder. The solution is instead to substitute \displaystyle u=x^2. If we write the integral as

\displaystyle \int\limits_0^1 x^3e^{x^2}\,dx = \int\limits_0^1 x^2e^{x^2}x\,dx

we see that the expression "\displaystyle x\,dx" can be replaced by \displaystyle du and the rest of the integrand contains only \displaystyle x in the form of \displaystyle x^2. The substitution gives

\displaystyle \begin{align} \int\limits_0^1 x^3e^{x^2}\,dx &= \int\limits_0^1 x^2e^{x^2}x\,dx\\[5pt] &= \left\{\begin{align} u &= x^2\\[5pt] du &= \bigl(x^2\bigr)'\,dx = 2x\,dx \end{align}\right\}\\[5pt] &= \int\limits_0^1 ue^u\tfrac{1}{2}\,du\\[5pt] &= \frac{1}{2}\int\limits_0^1 ue^u\,du\,\textrm{.} \end{align}

We can then calculate this integral by integration by parts, where we differentiate away the factor \displaystyle u,

\displaystyle \begin{align} \frac{1}{2}\int\limits_0^1 ue^u\,du &= \frac{1}{2}\Bigl[\ ue^u\ \Bigr]_0^1 - \frac{1}{2}\int\limits_0^1 1\cdot e^u\,du\\[5pt] &= \frac{1}{2}\bigl(1\cdot e^1-0\bigr) - \frac{1}{2}\Bigl[\ e^u\ \Bigr]_0^1\\[5pt] &= \frac{1}{2}e - \frac{1}{2}\bigl(e^1-e^0\bigr)\\[5pt] &= \frac{1}{2}e - \frac{1}{2}e + \frac{1}{2}\\[5pt] &= \frac{1}{2}\,\textrm{.} \end{align}