Solve ∫ 1-2/150+{left(65/309right)}^2 wrt x

$$\int{ 1- \frac{ 2 }{ 150 } + { \left( \frac{ 65 }{ 309 } \right) }^{ 2 } }d x$$

Evaluate

$$\frac{2460823x}{2387025}+С$$

Solution Steps

Reduce the fraction \frac{2}{150} to lowest terms by extracting and canceling out 2.

$$\int 1-\frac{1}{75}+\left(\frac{65}{309}\right)^{2}\mathrm{d}x$$

Convert 1 to fraction \frac{75}{75}.

$$\int \frac{75}{75}-\frac{1}{75}+\left(\frac{65}{309}\right)^{2}\mathrm{d}x$$

Since \frac{75}{75} and \frac{1}{75} have the same denominator, subtract them by subtracting their numerators.

$$\int \frac{75-1}{75}+\left(\frac{65}{309}\right)^{2}\mathrm{d}x$$

Subtract 1 from 75 to get 74.

$$\int \frac{74}{75}+\left(\frac{65}{309}\right)^{2}\mathrm{d}x$$

Calculate \frac{65}{309} to the power of 2 and get \frac{4225}{95481}.

$$\int \frac{74}{75}+\frac{4225}{95481}\mathrm{d}x$$

Least common multiple of 75 and 95481 is 2387025. Convert \frac{74}{75} and \frac{4225}{95481} to fractions with denominator 2387025.

$$\int \frac{2355198}{2387025}+\frac{105625}{2387025}\mathrm{d}x$$

Since \frac{2355198}{2387025} and \frac{105625}{2387025} have the same denominator, add them by adding their numerators.

$$\int \frac{2355198+105625}{2387025}\mathrm{d}x$$

Add 2355198 and 105625 to get 2460823.

$$\int \frac{2460823}{2387025}\mathrm{d}x$$

Find the integral of \frac{2460823}{2387025} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2460823x}{2387025}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{2460823x}{2387025}+С$$

Differentiate w.r.t. x

$$\frac{2460823}{2387025} = 1\frac{73798}{2387025} = 1.0309163079565569$$