Help on Integrals

**xmizzdelaneyx** · July 25th 2012, 08:07 PM

Can anybody help me with these two and explain it? I have tried a million times and can't get it right.

**Soroban** · July 25th 2012, 08:50 PM

Hello, xmizzdelaneyx!

$\text{Evaluate: }\:\int x\sqrt[4]{x+15}\,dx$

Let $\sqrt[4]{x+15} \:=\:u \quad\Rightarrow\quad x+15 \:=\:u^4 \quad\Rightarrow\quad x \:=\:u^4-15 \quad\Rightarrow\quad dx \:=\:4u^3\,du$

Substitute: . $\int(u^4-5)\cdot u \cdot (4u^3\,du) \;=\; 4\int(u^8 - 5u^4)\,du$

Can you finish it now?

$\text{Evaluate: }\:\int^3_{\frac{1}{7}}4x\ln(6x)\,dx$

By parts: . $\begin{Bmatrix}u &=& \ln(7x) && dv &=& 4x\,dx \\ du &=& \frac{dx}{x} && v &=& 2x^2 \end{Bmatrix}$

We have: . $2x^2\ln(7x) - \int(2x^2)\left(\frac{dx}{x}\right) \;=\;2x^2\ln(7x) - 2\int x\,dx$

. . . . . . . $=\;2x^2\ln(7x) - x^2\bigg]^3_{\frac{1}{7}}$

Therefore: . $\bigg[2\cdot9\cdot\ln(21) - 9\bigg] - \bigg[2\cdot\tfrac{1}{49}\cdot\ln(1) - \tfrac{1}{49}\bigg]$

. . . . . . $=\;\;18\ln(21) - 9 - 0 + \tfrac{1}{49} \;\;=\;\;18\ln(21) - \tfrac{440}{9}$

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