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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 x (because the original integral was expressed in x).

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

du=u(x)dx=(x2+3)dx=2xdx

which can be written as

xdx=21du.

The expression xdx is present as a factor in the integral, and so everything is there for the substitution u=x2+3,

(x2+3)5xdx=udu=x2+3=2xdx=u521du.

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

21u5du=216u6+C.

We write the answer expressed in x by substituting back u=x2+3,

(x2+3)5xdx=12(x2+3)6+C

where C is an arbitrary constant.

Note: It is possible to check the answer by differentiating 112(x2+3)6+C and seeing that we get back the integrand (x2+3)5x.