Evaluate each indefinite integral

Evaluate each indefinite integral 1) \int -2x\sqrt{9-x^2} dx 2) \int x^3(x^4+3)^2 dx 3) \int x\sqrt{x^2+2} dx 4) \int x^2 \cos(x^3) dx 5) \int \cos^2(4x)\sin(4x) dx 6) \int xe^{9-2x^2} dx 7) \int (2x^3-7) dx 8) \int (\frac{3}{x} + 2e^x + 5) dx 9) \int (\frac{1}{3x^2} – \frac{2x^{1/2}}{3}) dx 10) \int (\frac{3x^2+2}{x^3+2x}) dx

The Correct Answer and Explanation is:

Let’s evaluate each indefinite integral and provide a 300-word explanation afterward.

Solutions:

∫−2×9−x2 dx\int -2x\sqrt{9 – x^2} \, dx∫−2×9−x2​dx

Use substitution: u=9−x2u = 9 – x^2u=9−x2, then du=−2x dxdu = -2x \, dxdu=−2xdx.=∫u du=23u3/2+C=23(9−x2)3/2+C= \int \sqrt{u} \, du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}(9 – x^2)^{3/2} + C=∫u​du=32​u3/2+C=32​(9−x2)3/2+C

∫x3(x4+3)2 dx\int x^3(x^4 + 3)^2 \, dx∫x3(x4+3)2dx

Let u=x4+3u = x^4 + 3u=x4+3, so du=4x3dxdu = 4x^3 dxdu=4x3dx=14∫u2 du=112u3+C=112(x4+3)3+C= \frac{1}{4} \int u^2 \, du = \frac{1}{12} u^3 + C = \frac{1}{12}(x^4 + 3)^3 + C=41​∫u2du=121​u3+C=121​(x4+3)3+C

∫xx2+2 dx\int x\sqrt{x^2 + 2} \, dx∫xx2+2​dx

Use substitution: u=x2+2u = x^2 + 2u=x2+2, du=2xdxdu = 2x dxdu=2xdx=12∫u du=13u3/2+C=13(x2+2)3/2+C= \frac{1}{2} \int \sqrt{u} \, du = \frac{1}{3} u^{3/2} + C = \frac{1}{3}(x^2 + 2)^{3/2} + C=21​∫u​du=31​u3/2+C=31​(x2+2)3/2+C

∫x2cos⁡(x3) dx\int x^2 \cos(x^3) \, dx∫x2cos(x3)dx

Let u=x3u = x^3u=x3, du=3x2dxdu = 3x^2 dxdu=3x2dx=13∫cos⁡(u) du=13sin⁡(x3)+C= \frac{1}{3} \int \cos(u) \, du = \frac{1}{3} \sin(x^3) + C=31​∫cos(u)du=31​sin(x3)+C

∫cos⁡2(4x)sin⁡(4x) dx\int \cos^2(4x)\sin(4x) \, dx∫cos2(4x)sin(4x)dx

Let u=cos⁡(4x)u = \cos(4x)u=cos(4x), du=−4sin⁡(4x)dxdu = -4\sin(4x) dxdu=−4sin(4x)dx=−14∫u2 du=−112u3+C=−112cos⁡3(4x)+C= -\frac{1}{4} \int u^2 \, du = -\frac{1}{12} u^3 + C = -\frac{1}{12} \cos^3(4x) + C=−41​∫u2du=−121​u3+C=−121​cos3(4x)+C

∫xe9−2×2 dx\int xe^{9 – 2x^2} \, dx∫xe9−2x2dx

Let u=9−2x2u = 9 – 2x^2u=9−2×2, du=−4xdxdu = -4x dxdu=−4xdx=−14∫eu du=−14e9−2×2+C= -\frac{1}{4} \int e^u \, du = -\frac{1}{4} e^{9 – 2x^2} + C=−41​∫eudu=−41​e9−2×2+C

∫(2×3−7) dx=24×4−7x+C=12×4−7x+C\int (2x^3 – 7) \, dx = \frac{2}{4}x^4 – 7x + C = \frac{1}{2}x^4 – 7x + C∫(2×3−7)dx=42​x4−7x+C=21​x4−7x+C

∫(3x+2ex+5) dx=3ln⁡∣x∣+2ex+5x+C\int \left( \frac{3}{x} + 2e^x + 5 \right) \, dx = 3\ln|x| + 2e^x + 5x + C∫(x3​+2ex+5)dx=3ln∣x∣+2ex+5x+C

∫(13×2−2×1/23)dx=−13x−49×3/2+C\int \left( \frac{1}{3x^2} – \frac{2x^{1/2}}{3} \right) dx = -\frac{1}{3x} – \frac{4}{9}x^{3/2} + C∫(3×21​−32×1/2​)dx=−3×1​−94​x3/2+C

∫3×2+2×3+2x dx\int \frac{3x^2 + 2}{x^3 + 2x} \, dx∫x3+2x3x2+2​dx

Factor denominator: x(x2+2)x(x^2 + 2)x(x2+2)
Use partial fractions:3×2+2x(x2+2)=Ax+Bx+Cx2+2\frac{3x^2 + 2}{x(x^2 + 2)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 2}x(x2+2)3×2+2​=xA​+x2+2Bx+C​

Solving gives:
A=1,B=2,C=0A = 1, B = 2, C = 0A=1,B=2,C=0=∫(1x+2xx2+2)dx=ln⁡∣x∣+ln⁡(x2+2)+C= \int \left( \frac{1}{x} + \frac{2x}{x^2 + 2} \right) dx = \ln|x| + \ln(x^2 + 2) + C=∫(x1​+x2+22x​)dx=ln∣x∣+ln(x2+2)+C

Explanation

Solving indefinite integrals requires recognizing patterns and using techniques such as substitution, integration by parts, or algebraic manipulation. In the first few problems, substitution is the primary method. For example, when the integrand involves a product of a function and its derivative, substitution simplifies it into a standard integral. In problem 1, substituting u=9−x2u = 9 – x^2u=9−x2 makes the square root manageable. Similarly, in problems 2 through 4, expressions involving powers or square roots of polynomials benefit from substitution as well.

Problem 5 involves a trigonometric expression where recognizing the chain rule in reverse helps. We substitute u=cos⁡(4x)u = \cos(4x)u=cos(4x) and identify that the remaining factor is related to its derivative. This simplifies the integral into a polynomial in terms of cosine.

Problems 6 and 7 are straightforward—problem 6 again benefits from substitution, while problem 7 is a direct application of the power rule for integration.

In problems 8 and 9, the integrands are sums of simpler terms, allowing term-by-term integration. Logarithmic integration arises from ∫1x\int \frac{1}{x}∫x1​, and power rules apply to polynomial and radical terms.

Problem 10 is the most complex. It requires partial fraction decomposition to break the rational function into simpler pieces. Factoring the denominator and matching coefficients allows separating the integrand into terms whose antiderivatives are known—logarithmic in this case.

Overall, recognizing substitution opportunities, decomposing complex expressions, and recalling standard antiderivatives are crucial skills. Integration blends pattern recognition with algebraic and trigonometric techniques to reduce challenging expressions into solvable forms.