Solve ∫ (x^5-7)^2 wrt x=

$$\int ( x ^ { 5 } - 7 ) ^ { 2 } d x =$$

Evaluate

$$\frac{x^{11}}{11}-\frac{7x^{6}}{3}+49x+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{5}-7\right)^{2}.

$$\int \left(x^{5}\right)^{2}-14x^{5}+49\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 5 and 2 to get 10.

$$\int x^{10}-14x^{5}+49\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{10}\mathrm{d}x+\int -14x^{5}\mathrm{d}x+\int 49\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{10}\mathrm{d}x-14\int x^{5}\mathrm{d}x+\int 49\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{10}\mathrm{d}x with \frac{x^{11}}{11}.

$$\frac{x^{11}}{11}-14\int x^{5}\mathrm{d}x+\int 49\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply -14 times \frac{x^{6}}{6}.

$$\frac{x^{11}}{11}-\frac{7x^{6}}{3}+\int 49\mathrm{d}x$$

Find the integral of 49 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{11}}{11}-\frac{7x^{6}}{3}+49x$$

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.

$$49x-\frac{7x^{6}}{3}+\frac{x^{11}}{11}+С$$

Differentiate w.r.t. x

$$\left(x^{5}-7\right)^{2}$$