# How do you evaluate the integral of #int ln(1+x^3)dx#?

It is

Using integration by parts we have that

Now for the integral

#int [3x^3]/(1+x^3)dx=int (x-2)/(x^2-x+1)dx-int1/(x+1)+int 3dx= int (x-2)/(x^2-x+1)dx-lnabs(x+1)+3x#

Now for the integral

#int (x-2)/(x^2-x+1)dx=int 1/2[2x-1]/[x^2-x+1]dx+int [3/2]/[x^2-x+1]dx= 1/2*ln(x^2-x+1)-3/2int 1/[x^2-x+1]dx#

Now we have to calculate the integral as follows

#int 1/[x^2-x+1]dx=int 1/[(x-1/2)^2+3/4]dx= 4/3 int 1/[((x-1/2)/(sqrt3/2))^2+1]dx#

Now we set

Hence

Thus now the integral becomes

#4/3* int 1/[tan^2t+1]*(sqrt3/2)*sec^2t*dt=
2/{sqrt3]*tan^-1[(2x-1)/3]#

Finally we get that

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To evaluate the integral of ( \int \ln(1+x^3) , dx ), we can use integration by parts with ( u = \ln(1+x^3) ) and ( dv = dx ). Then, we differentiate ( u ) and integrate ( dv ) to find ( du ) and ( v ). After that, we apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

This gives us the integral:

[ \int \ln(1+x^3) , dx = x \ln(1+x^3) - \int x \left( \frac{3x^2}{1+x^3} \right) , dx ]

We can further simplify the remaining integral by making a substitution, letting ( t = 1 + x^3 ) and ( dt = 3x^2 , dx ), yielding:

[ \int x \left( \frac{3x^2}{1+x^3} \right) , dx = \int \frac{dt}{3} ]

Therefore,

[ \int \ln(1+x^3) , dx = x \ln(1+x^3) - \frac{1}{3} \int dt ]

[ = x \ln(1+x^3) - \frac{t}{3} + C ]

Where ( C ) is the constant of integration. Finally, we substitute back ( t = 1 + x^3 ):

[ = x \ln(1+x^3) - \frac{1+x^3}{3} + C ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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