#### To determine

**To evaluate:**

∫x3x2+1dx and verify using graphing function

#### Answer

∫x3x2+1dx=13x2+1 [x2-2]

#### Explanation

**1) Concept:**

i) Indefinite integration:

∫fx=F(x) means F'x=f(x)

ii) The Substitution rule:

If u=g(x) is a differentiable function whose range is an interval I and f is continuous on I, then

∫fgxg'(x) dx=∫f(u)du

**2) Formula:**

∫xndx=xn+1n+1+c

**3) Given:**

∫x3x2+1dx

**4) Calculation:**

Substitute, u=x2+1

Differentiating with respect to x,

dudx=2x

du2=x dx

∫x3x2+1dx=∫x2·xx2+1dx

So, x2=u-1

∫x2·xx2+1dx=∫u-1udu2

=12∫uu-1u du

=12∫u12-u-12 du

By using formula,

=1223u32-2u12

By plugging value of u,

=1223(x2+1)32-2(x2+1)12

=13(x2+1)32-(x2+1)12+C

Factor out 13(x2+1)12

=13(x2+1)12 [x2+1-3]+C

=13x2+1 [x2-2]+C

For, C=0

∫x3x2+1dx=13x2+1 [x2-2]

Now plot the graph of function f and its antiderivative F

From the graph, At point (0, 0) the graph of f changes from negative to positive, so it’s anti derivative will attain minima. To the left of zero f is negative so it’s anti derivative must be decreasing to the left of zero. To the right of zero since f is positive it’s anti derivative must increase. The computed function F thus has properties of anti derivative of f. Hence it is a reasonable answer.

**Conclusion:**

Therefore,

∫x3x2+1dx=13x2+1 [x2-2]