How can I prove the integral?

Question

Prove that $$ \int\frac{dx}{x(\log_e x)^{7/8}} = 8(\log_e x)^{1/8} $$

I am totally lost on this subject. Any help how to prove this is appreciated!

Answer 1

Hint: $\dfrac{d}{dx}(\log_e(x)) = \dfrac 1x$.

Putting $u = \log_e x \implies du = \frac 1x\,dx$.

That gives you:

$$ \int \frac{dx}{x(\log_e x)^{7/8}} = \int (\underbrace{\log_e x}_{u})^{-7/8}\cdot \underbrace{\left(\frac 1x \,dx \right)}_{du}=\int u^{-7/8} \,du$$

Can you take it from here (using the power rule)?

SPOILER (Scroll over to check your work!)

$$\int u^{-7/8}\,du = \frac{u^{1/8}}{1/8} + C = 8u^{1/8} + C = 8(\log_e x)^{1/8} + C$$

Answer 2

HINT:

If you have meant,$$\int\frac{dx}{x\{\log_e(e\cdot x)\}^{\dfrac78}}$$

As $\ln(ex)=\ln e+\ln x=1+\ln x$

Substitute this with $u$

---

If you have meant,$$\int\frac{dx}{x(\log_ex)^{\dfrac78}}$$

Substitute $\log_ex$ with $v$

Answer 3

Hint: set $t=\log_e x$, so $dt=\frac{1}{x}\,dx$

Answer 4

Firstly lets go with the convention of $ \log_ex=\ln x $, now $(\ln x)'=\frac {1}{x}$ so we have:

$\int (\frac {1}{x}(\ln x)^{-7/8})dx=\int(\ln x)^{-7/8})d(\ln x)=8(\ln x)^{1/8}+C$