Solve ∫ 10^-2(3-7x)^2(91+292x)^2 wrt x

$$\int 10 ^ { - 2 } ( 3 - 7 x ) ^ { 2 } ( 91 + 292 x ) ^ { 2 } d x$$

Evaluate

$$\frac{1044484x^{5}}{125}-\frac{122129x^{4}}{50}-\frac{1058903x^{3}}{300}+\frac{65247x^{2}}{100}+\frac{74529x}{100}+С$$

Solution Steps

Calculate 10 to the power of -2 and get \frac{1}{100}.

$$\int \frac{1}{100}\left(3-7x\right)^{2}\left(91+292x\right)^{2}\mathrm{d}x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-7x\right)^{2}.

$$\int \frac{1}{100}\left(9-42x+49x^{2}\right)\left(91+292x\right)^{2}\mathrm{d}x$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(91+292x\right)^{2}.

$$\int \frac{1}{100}\left(9-42x+49x^{2}\right)\left(8281+53144x+85264x^{2}\right)\mathrm{d}x$$

Use the distributive property to multiply \frac{1}{100} by 9-42x+49x^{2}.

$$\int \left(\frac{9}{100}-\frac{21}{50}x+\frac{49}{100}x^{2}\right)\left(8281+53144x+85264x^{2}\right)\mathrm{d}x$$

Use the distributive property to multiply \frac{9}{100}-\frac{21}{50}x+\frac{49}{100}x^{2} by 8281+53144x+85264x^{2} and combine like terms.

$$\int \frac{74529}{100}+\frac{65247}{50}x-\frac{1058903}{100}x^{2}-\frac{244258}{25}x^{3}+\frac{1044484}{25}x^{4}\mathrm{d}x$$

Integrate the sum term by term.

$$\int \frac{74529}{100}\mathrm{d}x+\int \frac{65247x}{50}\mathrm{d}x+\int -\frac{1058903x^{2}}{100}\mathrm{d}x+\int -\frac{244258x^{3}}{25}\mathrm{d}x+\int \frac{1044484x^{4}}{25}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int \frac{74529}{100}\mathrm{d}x+\frac{65247\int x\mathrm{d}x}{50}-\frac{1058903\int x^{2}\mathrm{d}x}{100}-\frac{244258\int x^{3}\mathrm{d}x}{25}+\frac{1044484\int x^{4}\mathrm{d}x}{25}$$

Find the integral of \frac{74529}{100} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{74529x}{100}+\frac{65247\int x\mathrm{d}x}{50}-\frac{1058903\int x^{2}\mathrm{d}x}{100}-\frac{244258\int x^{3}\mathrm{d}x}{25}+\frac{1044484\int x^{4}\mathrm{d}x}{25}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply \frac{65247}{50} times \frac{x^{2}}{2}.

$$\frac{74529x}{100}+\frac{65247x^{2}}{100}-\frac{1058903\int x^{2}\mathrm{d}x}{100}-\frac{244258\int x^{3}\mathrm{d}x}{25}+\frac{1044484\int x^{4}\mathrm{d}x}{25}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -\frac{1058903}{100} times \frac{x^{3}}{3}.

$$\frac{74529x}{100}+\frac{65247x^{2}}{100}-\frac{1058903x^{3}}{300}-\frac{244258\int x^{3}\mathrm{d}x}{25}+\frac{1044484\int x^{4}\mathrm{d}x}{25}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -\frac{244258}{25} times \frac{x^{4}}{4}.

$$\frac{74529x}{100}+\frac{65247x^{2}}{100}-\frac{1058903x^{3}}{300}-\frac{122129x^{4}}{50}+\frac{1044484\int x^{4}\mathrm{d}x}{25}$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply \frac{1044484}{25} times \frac{x^{5}}{5}.

$$\frac{74529x}{100}+\frac{65247x^{2}}{100}-\frac{1058903x^{3}}{300}-\frac{122129x^{4}}{50}+\frac{1044484x^{5}}{125}$$

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.

$$\frac{74529x}{100}+\frac{65247x^{2}}{100}-\frac{1058903x^{3}}{300}-\frac{122129x^{4}}{50}+\frac{1044484x^{5}}{125}+С$$

Differentiate w.r.t. x

$$\frac{\left(\left(7x-3\right)\left(292x+91\right)\right)^{2}}{100}$$