Solve ∫ 3^{2}*3^{5}:(5*3-7*2+2)^{6}+7]^2-(3*5^2+2^4)+(6-2^2)

$$\int 3 ^ { 2 } \times 3 ^ { 5 } : ( 5 \times 3 - 7 \times 2 + 2 ) ^ { 6 } + 7 ] ^ { 2 } - ( 3 \times 5 ^ { 2 } + 2 ^ { 4 } ) + ( 6 - 2 ^ { 2 } )$$

Evaluate

$$10x+С$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 5 to get 7.

$$\int \frac{3^{7}}{\left(5\times 3-7\times 2+2\right)^{6}}+7\mathrm{d}x$$

Calculate 3 to the power of 7 and get 2187.

$$\int \frac{2187}{\left(5\times 3-7\times 2+2\right)^{6}}+7\mathrm{d}x$$

Multiply 5 and 3 to get 15.

$$\int \frac{2187}{\left(15-7\times 2+2\right)^{6}}+7\mathrm{d}x$$

Multiply 7 and 2 to get 14.

$$\int \frac{2187}{\left(15-14+2\right)^{6}}+7\mathrm{d}x$$

Subtract 14 from 15 to get 1.

$$\int \frac{2187}{\left(1+2\right)^{6}}+7\mathrm{d}x$$

Add 1 and 2 to get 3.

$$\int \frac{2187}{3^{6}}+7\mathrm{d}x$$

Calculate 3 to the power of 6 and get 729.

$$\int \frac{2187}{729}+7\mathrm{d}x$$

Divide 2187 by 729 to get 3.

$$\int 3+7\mathrm{d}x$$

Add 3 and 7 to get 10.

$$\int 10\mathrm{d}x$$

Find the integral of 10 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$10x$$

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.

$$10x+С$$

Differentiate w.r.t. x

$$10$$