Solve ∫ (4x+3)^3 wrt x

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

Evaluate

$$\frac{\left(4x+3\right)^{4}}{16}+С$$

Solution Steps

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

$$\int 64x^{3}+144x^{2}+108x+27\mathrm{d}x$$

Integrate the sum term by term.

$$\int 64x^{3}\mathrm{d}x+\int 144x^{2}\mathrm{d}x+\int 108x\mathrm{d}x+\int 27\mathrm{d}x$$

Factor out the constant in each of the terms.

$$64\int x^{3}\mathrm{d}x+144\int x^{2}\mathrm{d}x+108\int x\mathrm{d}x+\int 27\mathrm{d}x$$

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

$$16x^{4}+144\int x^{2}\mathrm{d}x+108\int x\mathrm{d}x+\int 27\mathrm{d}x$$

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

$$16x^{4}+48x^{3}+108\int x\mathrm{d}x+\int 27\mathrm{d}x$$

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

$$16x^{4}+48x^{3}+54x^{2}+\int 27\mathrm{d}x$$

Find the integral of 27 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$16x^{4}+48x^{3}+54x^{2}+27x$$

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.

$$16x^{4}+48x^{3}+54x^{2}+27x+С$$

Differentiate w.r.t. x

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