Solve ∫ [(4x+2)^2+5]^2 wrt x

$$\int [ ( 4 x + 2 ) ^ { 2 } + 5 ] ^ { 2 } d x$$

Evaluate

$$\frac{256x^{5}}{5}+128x^{4}+\frac{544x^{3}}{3}+144x^{2}+81x+С$$

Solution Steps

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

$$\int \left(16x^{2}+16x+4+5\right)^{2}\mathrm{d}x$$

Add 4 and 5 to get 9.

$$\int \left(16x^{2}+16x+9\right)^{2}\mathrm{d}x$$

Square 16x^{2}+16x+9.

$$\int 256x^{4}+512x^{3}+544x^{2}+288x+81\mathrm{d}x$$

Integrate the sum term by term.

$$\int 256x^{4}\mathrm{d}x+\int 512x^{3}\mathrm{d}x+\int 544x^{2}\mathrm{d}x+\int 288x\mathrm{d}x+\int 81\mathrm{d}x$$

Factor out the constant in each of the terms.

$$256\int x^{4}\mathrm{d}x+512\int x^{3}\mathrm{d}x+544\int x^{2}\mathrm{d}x+288\int x\mathrm{d}x+\int 81\mathrm{d}x$$

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 256 times \frac{x^{5}}{5}.

$$\frac{256x^{5}}{5}+512\int x^{3}\mathrm{d}x+544\int x^{2}\mathrm{d}x+288\int x\mathrm{d}x+\int 81\mathrm{d}x$$

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

$$\frac{256x^{5}}{5}+128x^{4}+544\int x^{2}\mathrm{d}x+288\int x\mathrm{d}x+\int 81\mathrm{d}x$$

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

$$\frac{256x^{5}}{5}+128x^{4}+\frac{544x^{3}}{3}+288\int x\mathrm{d}x+\int 81\mathrm{d}x$$

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 288 times \frac{x^{2}}{2}.

$$\frac{256x^{5}}{5}+128x^{4}+\frac{544x^{3}}{3}+144x^{2}+\int 81\mathrm{d}x$$

Find the integral of 81 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{256x^{5}}{5}+128x^{4}+\frac{544x^{3}}{3}+144x^{2}+81x$$

Simplify.

$$81x+144x^{2}+\frac{544x^{3}}{3}+128x^{4}+\frac{256x^{5}}{5}$$

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.

$$81x+144x^{2}+\frac{544x^{3}}{3}+128x^{4}+\frac{256x^{5}}{5}+С$$

Differentiate w.r.t. x

$$\left(16x^{2}+16x+9\right)^{2}$$