The result of
∫(3x2−2x+7)73x−1dx
is ....
Explanation
To solve this integral, use the substitution method by letting u=3x2−2x+7
Determining the Substitution
Let u=3x2−2x+7, then its derivative is
dxdu=6x−2
du=2(3x−1)dx
From this, we obtain (3x−1)dx=21du
Transforming the Integral
Substitute into the integral
∫(3x2−2x+7)73x−1dx=∫u71⋅21du
=21∫u−7du
Integrating
21∫u−7du=21⋅−6u−6+C
=21⋅(−61)u−6+C
=−121u−6+C
Substituting Back
Substitute back u=3x2−2x+7
−121u−6+C=12(3x2−2x+7)6−1+C
Therefore, the result of ∫(3x2−2x+7)73x−1dx=12(3x2−2x+7)6−1+C