Let’s solve the integrals from the uploaded image using integration by inspection, providing detailed explanations for each case:

(a) $(4x - 7)^5$

We use substitution:

• Let $u = 4x - 7$, then $\frac{du}{dx} = 4 \implies dx = \frac{du}{4}$.

The integral becomes: $\int (4x - 7)^5 dx = \int u^5 \cdot \frac{1}{4} du = \frac{1}{4} \cdot \frac{u^6}{6} + C = \frac{(4x - 7)^6}{24} + C.$

(b) $\frac{1}{3x + 1}$

We use substitution:

• Let $u = 3x + 1$, so $\frac{du}{dx} = 3 \implies dx = \frac{du}{3}$.

The integral becomes: $\int \frac{1}{3x + 1} dx = \int \frac{1}{u} \cdot \frac{1}{3} du = \frac{1}{3} \ln|u| + C = \frac{1}{3} \ln|3x + 1| + C.$

(c) $\frac{1}{\sqrt{2x + 3}}$

We use substitution:

• Let $u = 2x + 3$, so $\frac{du}{dx} = 2 \implies dx = \frac{du}{2}$.

The integral becomes: $\int \frac{1}{\sqrt{2x + 3}} dx = \int \frac{1}{\sqrt{u}} \cdot \frac{1}{2} du = \frac{1}{2} \cdot 2\sqrt{u} + C = \sqrt{2x + 3} + C.$

(d) $x(1 + x^2)^4$

We use substitution:

• Let $u = 1 + x^2$, so $\frac{du}{dx} = 2x \implies x dx = \frac{1}{2} du$.

The integral becomes: $\int x(1 + x^2)^4 dx = \int u^4 \cdot \frac{1}{2} du = \frac{1}{2} \cdot \frac{u^5}{5} + C = \frac{(1 + x^2)^5}{10} + C.$

(e) $\frac{x^3}{(1 + x^4)^2}$

We use substitution:

• Let $u = 1 + x^4$, so $\frac{du}{dx} = 4x^3 \implies x^3 dx = \frac{1}{4} du$.

The integral becomes: $\int \frac{x^3}{(1 + x^4)^2} dx = \int \frac{1}{u^2} \cdot \frac{1}{4} du = \frac{1}{4} \cdot \left(-\frac{1}{u}\right) + C = -\frac{1}{4(1 + x^4)} + C.$

(f) $xe^{-x^2}$

We use substitution:

• Let $u = -x^2$, so $\frac{du}{dx} = -2x \implies x dx = -\frac{1}{2} du$.

The integral becomes: $\int xe^{-x^2} dx = \int e^u \cdot -\frac{1}{2} du = -\frac{1}{2} e^u + C = -\frac{1}{2} e^{-x^2} + C.$

(g) $\frac{1 + \ln x}{x}$

We split the integral into two parts: $\int \frac{1 + \ln x}{x} dx = \int \frac{1}{x} dx + \int \frac{\ln x}{x} dx.$

The first term: $\int \frac{1}{x} dx = \ln|x|.$

The second term (using substitution $u = \ln x$, so $du = \frac{1}{x} dx$): $\int \frac{\ln x}{x} dx = \int u du = \frac{u^2}{2} = \frac{(\ln x)^2}{2}.$

Thus: $\int \frac{1 + \ln x}{x} dx = \ln|x| + \frac{(\ln x)^2}{2} + C.$

(h) $\frac{x^2}{\sqrt{x^2 + 5}}$

We use substitution:

• Let $u = x^2 + 5$, so $\frac{du}{dx} = 2x \implies x dx = \frac{1}{2} du$.

The integral becomes: $\int \frac{x^2}{\sqrt{x^2 + 5}} dx = \int \frac{u - 5}{\sqrt{u}} \cdot \frac{1}{2} du = \frac{1}{2} \int \left(u^{\frac{1}{2}} - 5u^{-\frac{1}{2}}\right) du.$

Solve each term: $\int u^{\frac{1}{2}} du = \frac{2}{3} u^{\frac{3}{2}}, \quad \int u^{-\frac{1}{2}} du = 2u^{\frac{1}{2}}.$

Thus: $\frac{1}{2} \left(\frac{2}{3} u^{\frac{3}{2}} - 10u^{\frac{1}{2}}\right) + C = \frac{1}{3} (x^2 + 5)^{\frac{3}{2}} - 5(x^2 + 5)^{\frac{1}{2}} + C.$

Let me know if you'd like further clarifications!

5 Follow-Up Questions:

1. How is substitution determined in an integral problem?
2. Can you explain why the chain rule is essential in integration by substitution?
3. What are the main differences between definite and indefinite integrals?
4. How do you handle integrals that involve natural logarithmic terms like $\ln x$?
5. What happens when substitution doesn’t simplify the integral?

Tip:

Always check if the derivative of part of the integrand matches (or relates to) another part. This is the key to quickly identifying substitution.