Aus Online Mathematik Brückenkurs 2

For an indefinite integral, we do not need to take account of the limits of integration when substituting variables, but at the end, when the integral has been calculated, we do need to change back to the variable \displaystyle x (because the original integral was expressed in \displaystyle x).

If we start by looking at the integration element \displaystyle du, the relation between \displaystyle dx and \displaystyle du reads

\displaystyle du={u}'\left( x \right)\,dx=\left( x^{2}+3 \right)^{\prime }\,dx=2x\,dx,

which can be written as

\displaystyle x\,dx=\frac{1}{2}\,du

The expression \displaystyle x\,dx is present as a factor in the integral, and so everything is there for the substitution \displaystyle u=x^{2}+3

\displaystyle \int{\left( x^{2}+3 \right)}^{5}x\,dx=\left\{ \begin{matrix} u=x^{2}+3 \\ du=2x\,dx \\ \end{matrix} \right\}=\int{u^{5}}\centerdot \frac{1}{2}\,du

The result on the right-hand side is a standard integral, which we integrate directly,

\displaystyle \frac{1}{2}\int{u^{5}}\,du=\frac{1}{2}\centerdot \frac{u^{6}}{6}+C

We write the answer expressed in \displaystyle x\text{ } by substituting back \displaystyle u=x^{2}+3,

\displaystyle \int{\left( x^{2}+3 \right)}^{5}x\,dx=\frac{\left( x^{2}+3 \right)^{6}}{12}+C

where \displaystyle C is an arbitrary constant.

NOTE: it is possible to check the answer by differentiating \displaystyle \frac{1}{12}\left( x^{2}+3 \right)^{6}+C and seeing that we get back the integrand \displaystyle \left( x^{2}+3 \right)^{5}x.