Solve ∫ (5x^3+4)^2(15x^2+4) wrt x

$$\int ( 5 x ^ { 3 } + 4 ) ^ { 2 } ( 15 x ^ { 2 } + 4 ) d x$$

Evaluate

$$\frac{125x^{9}}{3}+\frac{100x^{7}}{7}+100x^{6}+40x^{4}+80x^{3}+64x+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x^{3}+4\right)^{2}.

$$\int \left(25\left(x^{3}\right)^{2}+40x^{3}+16\right)\left(15x^{2}+4\right)\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$\int \left(25x^{6}+40x^{3}+16\right)\left(15x^{2}+4\right)\mathrm{d}x$$

Use the distributive property to multiply 25x^{6}+40x^{3}+16 by 15x^{2}+4.

$$\int 375x^{8}+100x^{6}+600x^{5}+160x^{3}+240x^{2}+64\mathrm{d}x$$

Integrate the sum term by term.

$$\int 375x^{8}\mathrm{d}x+\int 100x^{6}\mathrm{d}x+\int 600x^{5}\mathrm{d}x+\int 160x^{3}\mathrm{d}x+\int 240x^{2}\mathrm{d}x+\int 64\mathrm{d}x$$

Factor out the constant in each of the terms.

$$375\int x^{8}\mathrm{d}x+100\int x^{6}\mathrm{d}x+600\int x^{5}\mathrm{d}x+160\int x^{3}\mathrm{d}x+240\int x^{2}\mathrm{d}x+\int 64\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{8}\mathrm{d}x with \frac{x^{9}}{9}. Multiply 375 times \frac{x^{9}}{9}.

$$\frac{125x^{9}}{3}+100\int x^{6}\mathrm{d}x+600\int x^{5}\mathrm{d}x+160\int x^{3}\mathrm{d}x+240\int x^{2}\mathrm{d}x+\int 64\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}. Multiply 100 times \frac{x^{7}}{7}.

$$\frac{125x^{9}}{3}+\frac{100x^{7}}{7}+600\int x^{5}\mathrm{d}x+160\int x^{3}\mathrm{d}x+240\int x^{2}\mathrm{d}x+\int 64\mathrm{d}x$$

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 600 times \frac{x^{6}}{6}.

$$\frac{125x^{9}}{3}+\frac{100x^{7}}{7}+100x^{6}+160\int x^{3}\mathrm{d}x+240\int x^{2}\mathrm{d}x+\int 64\mathrm{d}x$$

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 160 times \frac{x^{4}}{4}.

$$\frac{125x^{9}}{3}+\frac{100x^{7}}{7}+100x^{6}+40x^{4}+240\int x^{2}\mathrm{d}x+\int 64\mathrm{d}x$$

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 240 times \frac{x^{3}}{3}.

$$\frac{125x^{9}}{3}+\frac{100x^{7}}{7}+100x^{6}+40x^{4}+80x^{3}+\int 64\mathrm{d}x$$

Find the integral of 64 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{125x^{9}}{3}+\frac{100x^{7}}{7}+100x^{6}+40x^{4}+80x^{3}+64x$$

Simplify.

$$80x^{3}+64x+100x^{6}+40x^{4}+\frac{125x^{9}}{3}+\frac{100x^{7}}{7}$$

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.

$$80x^{3}+64x+100x^{6}+40x^{4}+\frac{125x^{9}}{3}+\frac{100x^{7}}{7}+С$$

Differentiate w.r.t. x

$$\left(15x^{2}+4\right)\left(5x^{3}+4\right)^{2}$$