Integrate the following functions: 1. integral (16x)/(8x2 + 2) dx 2. integral (x2)/(x3 + 1) dx

Question

ScanSolve AI Tutor · Accepted Answer

Here are the step-by-step solutions for the first two questions.

1. $\int \frac{16x}{8x^2 + 2}\,dx$

Use u-substitution.
Let $u = 8x^2 + 2$ .
Differentiate $u$ with respect to $x$ to find $du$ .
$du = \frac{d}{dx}(8x^2 + 2)\,dx = (16x)\,dx$
Substitute $u$ and $du$ into the integral.
$\int \frac{16x}{8x^2 + 2}\,dx = \int \frac{1}{u}\,du$
Integrate with respect to $u$ .
$\int \frac{1}{u}\,du = \ln|u| + C$
Substitute back $u = 8x^2 + 2$ .
$\ln|8x^2 + 2| + C$
Since $8x^2 + 2$ is always positive, the absolute value is not strictly necessary.
The final answer is $\boxed{\ln(8x^2 + 2) + C}$ .

2. $\int \frac{x^2}{x^3 + 1}\,dx$
Use u-substitution.
Let $u = x^3 + 1$ .
Differentiate $u$ with respect to $x$ to find $du$ .
$du = \frac{d}{dx}(x^3 + 1)\,dx = (3x^2)\,dx$
Rearrange to solve for $x^2\,dx$ .
$x^2\,dx = \frac{1}{3}\,du$
Substitute $u$ and $x^2\,dx$ into the integral.
$\int \frac{x^2}{x^3 + 1}\,dx = \int \frac{1}{u} \cdot \frac{1}{3}\,du = \frac{1}{3} \int \frac{1}{u}\,du$
Integrate with respect to $u$ .
$\frac{1}{3} \int \frac{1}{u}\,du = \frac{1}{3} \ln|u| + C$
Substitute back $u = x^3 + 1$ .
$\frac{1}{3} \ln|x^3 + 1| + C$
The final answer is $\boxed{\frac{1}{3} \ln|x^3 + 1| + C}$ .

Integrate the following functions: 1. integral (16x)/(8x2 + 2)\,dx 2. integral (x2)/(x3 + 1)\,dx