Exercises

Approach: Split the integrand into terms and apply the power rule to each term separately, then collect the results and restore the constant of integration.

Steps:

1. By linearity, $\int (3x^2 - 2x + 1)\,dx = 3\int x^2\,dx - 2\int x\,dx + \int 1\,dx$.
2. Integrate each term with the power rule: $3 \cdot \frac{x^3}{3} = x^3$, $-2 \cdot \frac{x^2}{2} = -x^2$, and $\int 1\,dx = x$.
3. Collect the terms and add the constant: $x^3 - x^2 + x + C$.
4. Check by differentiating: $\frac{d}{dx}(x^3 - x^2 + x + C) = 3x^2 - 2x + 1$.

Answer: $x^3 - x^2 + x + C$.

Approach: Rewrite the integrand as a negative power and apply the power rule, remembering that the exponent $-2$ is not the excluded value $-1$.

Steps:

1. Rewrite: $\frac{1}{x^2} = x^{-2}$, so the exponent is $n = -2$.
2. Apply the rule: $\int x^{-2}\,dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C$.
3. Check by differentiating: $\frac{d}{dx}\left(-\frac{1}{x}\right) = \frac{1}{x^2} = x^{-2}$.

Answer: $-\frac{1}{x} + C$.

Approach: Substitute $u = 3x$ so that the integrand becomes $e^u$ and the chain-rule factor $\frac{1}{3}$ appears, then integrate the exponential.

Steps:

1. Substitute $u = 3x$, so $du = 3\,dx$ and $dx = \frac{1}{3}du$.
2. Integrate: $\int e^{3x}\,dx = \frac{1}{3}\int e^u\,du = \frac{1}{3}e^u + C$.
3. Back-substitute: $\frac{1}{3}e^u + C = \frac{1}{3}e^{3x} + C$.
4. Check by differentiating: $\frac{d}{dx}\left(\frac{1}{3}e^{3x}\right) = \frac{1}{3}\cdot 3e^{3x} = e^{3x}$.

Answer: $\frac{1}{3}e^{3x} + C$.

Approach: Use the basic formula $\int \sin x\,dx = -\cos x + C$, which follows from the derivative $\frac{d}{dx}(-\cos x) = \sin x$. The minus sign comes from the direction of differentiation.

Steps:

1. Recall that $(\cos x)' = -\sin x$, so $-\cos x$ is an antiderivative of $\sin x$.
2. Add the constant of integration, since the indefinite integral denotes a whole family: $-\cos x + C$.
3. Check by differentiating: $\frac{d}{dx}(-\cos x + C) = \sin x$.

Answer: $-\cos x + C$.

Approach: The numerator $x$ is half the derivative of $x^2$, so substitute $u = x^2$ and the integral becomes an exponential one.

Steps:

1. Substitute $u = x^2$, so $du = 2x\,dx$ and therefore $x\,dx = \frac{1}{2}du$.
2. Integrate: $\int x e^{x^2}\,dx = \frac{1}{2}\int e^u\,du = \frac{1}{2}e^u + C$.
3. Back-substitute: $\frac{1}{2}e^u + C = \frac{1}{2}e^{x^2} + C$.
4. Check by differentiating: $\frac{d}{dx}\left(\frac{1}{2}e^{x^2}\right) = \frac{1}{2}\cdot 2x\,e^{x^2} = x e^{x^2}$.

Answer: $\frac{1}{2}e^{x^2} + C$.

Approach: The power-reduction identity $\sin^2 x = \frac{1 - \cos 2x}{2}$ rewrites the square in terms of terms that can be integrated directly.

Steps:

1. Apply the identity: $\sin^2 x = \frac{1 - \cos 2x}{2}$.
2. Integrate term by term: $\int \sin^2 x\,dx = \frac{1}{2}\int (1 - \cos 2x)\,dx = \frac{1}{2}x - \frac{1}{2}\cdot\frac{1}{2}\sin 2x + C = \frac{x}{2} - \frac{1}{4}\sin 2x + C$.
3. Check by differentiating: $\frac{d}{dx}\left(\frac{x}{2} - \frac{1}{4}\sin 2x\right) = \frac{1}{2} - \frac{1}{2}\cos 2x = \frac{1 - \cos 2x}{2} = \sin^2 x$.

Answer: $\frac{x}{2} - \frac{1}{4}\sin 2x + C$.

Approach: The denominator $1 + x^2$ is the signature of the inverse tangent: $\frac{d}{dx}\arctan x = \frac{1}{1+x^2}$.

Steps:

1. Recognize the pattern: the integrand is the derivative of $\arctan x$.
2. Integrate: $\int \frac{1}{x^2+1}\,dx = \arctan x + C$, with the constant restored because the indefinite integral is a family.
3. Check by differentiating: $\frac{d}{dx}(\arctan x + C) = \frac{1}{1+x^2}$.

Answer: $\arctan x + C$.

Approach: The denominator factors as $x^2 - 1 = (x-1)(x+1)$, so decompose the integrand into partial fractions and integrate each simple term.

Steps:

1. Decompose: $\frac{1}{x^2-1} = \frac{1}{(x-1)(x+1)} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)$.
2. Integrate term by term: $\int \frac{1}{x^2-1}\,dx = \frac{1}{2}\left(\ln\lvert x-1\rvert - \ln\lvert x+1\rvert\right) + C$.
3. Combine the logarithms: $\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C$.
4. Check by differentiating: $\frac{d}{dx}\left[\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|\right] = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right) = \frac{1}{x^2-1}$.

Answer: $\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C$, valid on any interval that does not contain $x = \pm 1$.

Approach: This is the power rule applied to four separate terms: split the integrand by linearity, integrate each piece, and collect the results.

Steps:

1. By linearity, $\int (x^3 + 2x^2 + x + 1)\,dx = \int x^3\,dx + 2\int x^2\,dx + \int x\,dx + \int 1\,dx$.
2. Integrate each term with the power rule: $\int x^3\,dx = \frac{x^4}{4}$, $2\int x^2\,dx = 2\cdot\frac{x^3}{3} = \frac{2x^3}{3}$, $\int x\,dx = \frac{x^2}{2}$, and $\int 1\,dx = x$.
3. Collect the terms and add the constant: $\frac{x^4}{4} + \frac{2x^3}{3} + \frac{x^2}{2} + x + C$.
4. Check by differentiating term by term: $\frac{d}{dx}\left(\frac{x^4}{4}\right) = x^3$, $\frac{d}{dx}\left(\frac{2x^3}{3}\right) = 2x^2$, $\frac{d}{dx}\left(\frac{x^2}{2}\right) = x$, and $\frac{d}{dx}(x) = 1$, so the derivative is $x^3 + 2x^2 + x + 1$, exactly the integrand.

Answer: $\frac{x^4}{4} + \frac{2x^3}{3} + \frac{x^2}{2} + x + C$.

Approach: Use the power-reduction identity $\cos^2 x = \frac{1 + \cos 2x}{2}$, which turns the square into a constant plus a cosine that integrates directly.

Steps:

1. Apply the identity: $\cos^2 x = \frac{1 + \cos 2x}{2}$.
2. Integrate term by term: $\int \cos^2 x\,dx = \frac{1}{2}\int (1 + \cos 2x)\,dx = \frac{1}{2}\left(x + \frac{1}{2}\sin 2x\right) + C = \frac{x}{2} + \frac{1}{4}\sin 2x + C$.
3. Check by differentiating: $\frac{d}{dx}\left(\frac{x}{2} + \frac{1}{4}\sin 2x\right) = \frac{1}{2} + \frac{1}{2}\cos 2x = \frac{1 + \cos 2x}{2} = \cos^2 x$.

Answer: $\frac{x}{2} + \frac{1}{4}\sin 2x + C$.

Approach: The numerator $x$ is half the derivative of $x^2 + 1$, so substitute $u = x^2 + 1$ and the integral becomes a plain logarithmic one.

Steps:

1. Substitute $u = x^2 + 1$, so $du = 2x\,dx$ and therefore $x\,dx = \frac{1}{2}du$.
2. Integrate: $\int \frac{x}{x^2 + 1}\,dx = \frac{1}{2}\int \frac{1}{u}\,du = \frac{1}{2}\ln\lvert u\rvert + C$.
3. Back-substitute: $\frac{1}{2}\ln\lvert x^2 + 1\rvert + C = \frac{1}{2}\ln(x^2 + 1) + C$, where the absolute value may be dropped because $x^2 + 1 \geq 1 > 0$ for every real $x$.
4. Check by differentiating: $\frac{d}{dx}\left(\frac{1}{2}\ln(x^2 + 1)\right) = \frac{1}{2}\cdot\frac{2x}{x^2 + 1} = \frac{x}{x^2 + 1}$.

Answer: $\frac{1}{2}\ln(x^2 + 1) + C$.

Approach: Instead of integrating, differentiate the proposed antiderivative and compare the result with the given integrand.

Steps:

1. Differentiate the right-hand side: $\frac{d}{dx}\left(e^x - \cos x + C\right) = e^x + \sin x$.
2. Compare the two expressions:
   • Integrand: $e^x + \sin x$
   • Derivative of the claim: $e^x + \sin x$
3. The derivative reproduces the integrand exactly, so the claim is correct; the only freedom left is the arbitrary constant $C$, which the indefinite integral must carry.

Answer: The claimed antiderivative is correct: $\int (e^x + \sin x)\,dx = e^x - \cos x + C$.

Approach: The integrand is the derivative of the inverse sine: $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}$, so the antiderivative is $\arcsin x$.

Steps:

1. Recognize the pattern: for $\lvert x\rvert < 1$, $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}$.
2. Integrate: $\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C$, with the constant restored because the indefinite integral is a family.
3. Check by differentiating with the chain rule: $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - (\sin(\arcsin x))^2}}\cdot(\sin(\arcsin x))' = \frac{1}{\sqrt{1-x^2}}$.
4. Note the domain: the integrand is real only for $\lvert x\rvert < 1$, so the antiderivative is valid on that interval.

Answer: $\arcsin x + C$, valid for $\lvert x\rvert < 1$.

Approach: Rewrite $\tan x$ as $\frac{\sin x}{\cos x}$ and then substitute $u = \cos x$, so that the numerator $dx$ factor cancels and a logarithmic integral remains.

Steps:

1. Rewrite: $\tan x = \frac{\sin x}{\cos x}$.
2. Substitute $u = \cos x$, so $du = -\sin x\,dx$ and therefore $\sin x\,dx = -du$.
3. Integrate: $\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx = \int \frac{1}{u}\cdot(-du) = -\int \frac{1}{u}\,du = -\ln\lvert u\rvert + C$.
4. Back-substitute: $-\ln\lvert \cos x\rvert + C$.
5. Check by differentiating: $\frac{d}{dx}\left(-\ln\lvert \cos x\rvert\right) = -\frac{1}{\cos x}\cdot(-\sin x) = \frac{\sin x}{\cos x} = \tan x$.

Answer: $-\ln\lvert \cos x\rvert + C$, valid on any interval that does not contain a zero of $\cos x$.

Approach: Complete the square in the denominator to get $(x + 1)^2$, then substitute $u = x + 1$ so that the integral becomes a power integral.

Steps:

1. Complete the square: $x^2 + 2x + 1 = (x + 1)^2$, so $\int \frac{1}{x^2 + 2x + 1}\,dx = \int (x + 1)^{-2}\,dx$.
2. Substitute $u = x + 1$, so $du = dx$.
3. Integrate with the power rule: $\int u^{-2}\,du = \frac{u^{-1}}{-1} + C = -\frac{1}{u} + C$.
4. Back-substitute: $-\frac{1}{x + 1} + C$.
5. Check by differentiating: $\frac{d}{dx}\left(-\frac{1}{x + 1}\right) = \frac{1}{(x + 1)^2} = \frac{1}{x^2 + 2x + 1}$.

Answer: $-\frac{1}{x + 1} + C$, valid on any interval that does not contain $x = -1$.