How do you evaluate the definite integral int x/(x+1)∫xx+1 from [0,1][0,1]?

3 Answers
Gió
Nov 16, 2016

I found: 1-ln(2)1−ln(2)

Explanation:

We can solve first the indefinite integral by manipulating a bit the integrand:
int(x/(x+1))dx=-int(1/(x+1)-1)dx=-[int1/(x+1)dx-intdx]=-[ln(x+1)-x]∫(xx+1)dx=−∫(1x+1−1)dx=−[∫1x+1dx−∫dx]=−[ln(x+1)−x]
evaluated between x=0x=0 and x=1x=1:

=-[ln(2)-1-ln(1)+0]=1-ln(2)=−[ln(2)−1−ln(1)+0]=1−ln(2)

Konstantinos Michailidis
Nov 16, 2016

We have that

int (x+1-1)/(x+1)*dx=int (1-1/(x+1))dx=x-ln(x+1)+c∫x+1−1x+1⋅dx=∫(1−1x+1)dx=x−ln(x+1)+c

Hence

int_0^1 x/(x+1)dx=[x-ln(x+1)]_0^1=1-ln2=0.30685∫10xx+1dx=[x−ln(x+1)]10=1−ln2=0.30685

Narad T.
Nov 16, 2016

The answer is =(1-ln2)=(1−ln2)

Explanation:

We do this by substitution, u=x+1u=x+1

Then du=dxdu=dx

So, int(xdx)/(x+1)=int((u-1)du)/u∫xdxx+1=∫(u−1)duu

=int(1-1/u)du=u-lnu +C=∫(1−1u)du=u−lnu+C

=(x+1)-ln(x+1)+C=(x+1)−ln(x+1)+C

Therefore,

int_0^1(xdx)/(x+1)=[(x+1)-ln(x+1)] _0^1∫10xdxx+1=[(x+1)−ln(x+1)]10

=2-ln2-1-0=(1-ln2)=2−ln2−1−0=(1−ln2)

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