Thread: Help on Integrals

1. Help on Integrals

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

2. Re: Help on Integrals

Hello, xmizzdelaneyx!

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

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

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

Can you finish it now?

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

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

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

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

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

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