Evaluate the integral: I = integral (x²+ sqrt(1+x))/(7th
Evaluate the integral: $$ I = \int \frac{x^2+\sqrt{1+x}}{\sqrt[7]{1+x}}\,dx. $$
Explanation of Method
Splitting the integrand into two terms and substituting $u=(1+x)^{1/14}$ (the least common denominator of the exponents $7$ and $2$) rationalizes both pieces into pure powers of $u$.
- Split the integrand into two terms.
- Substitute $u=(1+x)^{1/14}$.
- Integrate each resulting power of $u$.
- Back-substitute.
Step 1 · Split the integrand
$$ \frac{x^2+\sqrt{1+x}}{\sqrt[7]{1+x}} = \frac{x^2}{(1+x)^{1/7}} + (1+x)^{5/14}. $$
Step 2 · Substitute
Let $u=(1+x)^{1/14}$, so $x=u^{14}-1$, $dx=14u^{13}\,du$, $(1+x)^{1/7}=u^2$, $(1+x)^{5/14}=u^5$:
$$ \int\frac{x^2}{(1+x)^{1/7}}\,dx = 14\int u^{11}\left(u^{14}-1\right)^2 du = 14\int\left(u^{39}-2u^{25}+u^{11}\right)du, $$
$$ \int(1+x)^{5/14}\,dx = 14\int u^{18}\,du. $$
Step 3 · Integrate each power
$$ 14\int\left(u^{39}-2u^{25}+u^{11}\right)du = \frac{7u^{40}}{20}-\frac{14u^{26}}{13}+\frac{7u^{12}}6, \qquad 14\int u^{18}\,du = \frac{14u^{19}}{19}. $$
Step 4 · Combine and back-substitute
With $u^{40}=(1+x)^{20/7}$, $u^{26}=(1+x)^{13/7}$, $u^{12}=(1+x)^{6/7}$, $u^{19}=(1+x)^{19/14}$:
$$ I = \frac{7(1+x)^{20/7}}{20} - \frac{14(1+x)^{13/7}}{13} + \frac{7(1+x)^{6/7}}6 + \frac{14(1+x)^{19/14}}{19} + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $x^2/(1+x)^{1/7}$ | $\tfrac{7u^{40}}{20}-\tfrac{14u^{26}}{13}+\tfrac{7u^{12}}6$ |
| $(1+x)^{5/14}$ | $\tfrac{14u^{19}}{19}$ |
Additional Notes
- The domain requires $x\geq-1$.
- The result was verified numerically at $x=0$ (so $u=1$): the chain-rule derivative $14u^{11}(u^{14}-1)^2+14u^{18}$ times $du/dx$ evaluates to $1$, matching the integrand $(0+1)/1=1$ there.
Result
The solution to the integral is: