Integration techniques

Answer: $\ln\left(\frac{7}{2}\right)$

$\int \dfrac{1}{x + 2}dx = \ln\left|x + 2\right|\mathclose{.}$ From $0$ to $5\mathclose{:}$ $\ln(7) - \ln(2) = \ln\left(\frac{7}{2}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used only the upper limit → $\ln\left(7\right)$
• Differentiated instead, using $-\dfrac{1}{(x + a)^2}$ → $\dfrac{45}{196}$

Answer: $1$

By parts with $u = x$ and $dv = \sin(x)\,dx\mathclose{,}$ so $v = -\cos(x)\mathclose{:}$ $\int x\sin(x)\,dx = x \cdot \left(-\cos(x)\right) - \int \left(-\cos(x)\right)\,dx = -x\cos(x) + \sin(x)\mathclose{.}$ From $0$ to $\frac{\pi}{2}$ that is $1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the second term of integration by parts → $0$
• Got the sign of the second term wrong → $-1$

Answer: $-\dfrac{e^{\pi} + 1}{2}$

By parts twice the integral comes back: call it $I\mathclose{,}$ and the two steps give $I = (\text{known terms}) - I\mathclose{,}$ so $2I$ is known. The antiderivative is $\dfrac{e^{x}\left(\sin(x) + \cos(x)\right)}{2}\mathclose{,}$ and from $0$ to $\pi$ that is $-\dfrac{e^{\pi} + 1}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to divide by 2 once the integral came back → $-e^{\pi} - 1$
• Used only the upper limit → $-\dfrac{e^{\pi}}{2}$

Answer: $\dfrac{1}{2}\ln\left(\frac{7}{3}\right)$

The top is half the derivative of the bottom, so $\int \dfrac{x}{x^{2} + 3}dx = \dfrac{1}{2}\ln\left(x^{2} + 3\right)\mathclose{.}$ From $0$ to $2\mathclose{:}$ $\dfrac{1}{2}\ln(7) - \dfrac{1}{2}\ln(3) = \dfrac{1}{2}\ln\left(\frac{7}{3}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the factor $\frac{1}{2}$ → $\ln\left(\frac{7}{3}\right)$
• Used only the upper limit → $\dfrac{1}{2}\ln\left(7\right)$

Answer: $\dfrac{e^{2} + 1}{4}$

By parts with $u = x$ and $dv = e^{2x}dx\mathclose{:}$ $\int xe^{2x}dx = \dfrac{xe^{2x}}{2} - \int \dfrac{e^{2x}}{2}dx = \dfrac{xe^{2x}}{2} - \dfrac{e^{2x}}{4}\mathclose{.}$ From $0$ to $1$ that is $\dfrac{e^{2} + 1}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the second term of integration by parts → $\dfrac{e^{2}}{2}$
• Got the sign of the second term wrong → $\dfrac{3e^{2} - 1}{4}$

Answer: $e - 2$

By parts twice. With $u = x^2\mathclose{:}$ $\int x^2e^x dx = x^2e^x - \int 2xe^x dx\mathclose{;}$ with $u = 2x$ again: $\int 2xe^x dx = 2xe^x - 2e^x\mathclose{.}$ So the antiderivative is $e^x(x^2 - 2x + 2)\mathclose{,}$ and from $0$ to $1$ that is $e - 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stopped after one integration by parts, dropping the integral left over → $e$
• Added each integral instead of subtracting it → $5e - 2$

Answer: $-2 + 8\ln(3)$

Divide first: $\dfrac{x^{2} - x + 6}{x + 1} = x - 2 + \dfrac{8}{x + 1}\mathclose{.}$ So the integral is $\left[\dfrac{x^{2}}{2} - 2 x + 8\ln(x + 1)\right]_{0}^{2} = -2 + 8\ln(3)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Dropped the remainder after dividing → $-2$
• Divided only the leading terms, keeping just $x$ → $2$

Answer: $156$

Put $u = x^{2} + 1\mathclose{,}$ so $du = 2x\,dx\mathclose{,}$ and the limits become $u = 1$ and $u = 5\mathclose{:}$ the integral is $\displaystyle\int_{1}^{5} u^{3}\,du = \left[\dfrac{u^{4}}{4}\right]_{1}^{5} = 156\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to divide by the new power $4$ → $624$
• Kept the $x$-limits $0$ and $2$ for $u$ → $4$

Answer: $\pi$

Use the double-angle formula: $\cos^{2}(x) = \dfrac{1 + \cos(2x)}{2}\mathclose{.}$ Its antiderivative is $\dfrac{x}{2} + \dfrac{\sin(2x)}{4}\mathclose{,}$ and $\sin(2x)$ is $0$ at both limits, so the integral is $\dfrac{1}{2} \cdot 2\pi = \pi\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the $\frac{1}{2}$ of the double-angle formula → $2\pi$
• Integrated $\cos^2(x)$ like a power, as $\frac{1}{3}\cos^3(x)$ → $0$

Answer: $\dfrac{1}{3}\ln\left(2\right)$

Split into partial fractions: $\dfrac{1}{\left(x - 1\right)\left(x + 2\right)} = \dfrac{1}{3}\left(\dfrac{1}{x - 1} - \dfrac{1}{x + 2}\right)\mathclose{.}$ The antiderivative is $\dfrac{1}{3}\ln\left|\dfrac{x - 1}{x + 2}\right|\mathclose{;}$ from $2$ to $4$ that is $\dfrac{1}{3}\ln\left(2\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot the factor $\dfrac{1}{3}$ from the partial fractions → $\ln\left(2\right)$
• Added the two logarithms instead of subtracting them → $\dfrac{1}{3}\ln\left(\frac{9}{2}\right)$

Answer: $\ln\left(\frac{7}{5}\right)$

$\int \dfrac{1}{x + 5}dx = \ln\left|x + 5\right|\mathclose{.}$ From $0$ to $2\mathclose{:}$ $\ln(7) - \ln(5) = \ln\left(\frac{7}{5}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used only the upper limit → $\ln\left(7\right)$
• Differentiated instead, using $-\dfrac{1}{(x + a)^2}$ → $\dfrac{24}{1225}$

Answer: $\frac{\pi}{2} - 1$

By parts with $u = x$ and $dv = \cos(x)\,dx\mathclose{,}$ so $v = \sin(x)\mathclose{:}$ $\int x\cos(x)\,dx = x \cdot \sin(x) - \int \sin(x)\,dx = x\sin(x) + \cos(x)\mathclose{.}$ From $0$ to $\frac{\pi}{2}$ that is $\frac{\pi}{2} - 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the second term of integration by parts → $\frac{\pi}{2}$
• Got the sign of the second term wrong → $\frac{\pi}{2} + 1$

Answer: $\dfrac{e^{\frac{\pi}{2}} + 1}{2}$

By parts twice the integral comes back: call it $I\mathclose{,}$ and the two steps give $I = (\text{known terms}) - I\mathclose{,}$ so $2I$ is known. The antiderivative is $\dfrac{e^{x}\left(\sin(x) - \cos(x)\right)}{2}\mathclose{,}$ and from $0$ to $\frac{\pi}{2}$ that is $\dfrac{e^{\frac{\pi}{2}} + 1}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to divide by 2 once the integral came back → $e^{\frac{\pi}{2}} + 1$
• Used only the upper limit → $\dfrac{e^{\frac{\pi}{2}}}{2}$

Answer: $\dfrac{1}{2}\ln\left(\frac{13}{4}\right)$

The top is half the derivative of the bottom, so $\int \dfrac{x}{x^{2} + 4}dx = \dfrac{1}{2}\ln\left(x^{2} + 4\right)\mathclose{.}$ From $0$ to $3\mathclose{:}$ $\dfrac{1}{2}\ln(13) - \dfrac{1}{2}\ln(4) = \dfrac{1}{2}\ln\left(\frac{13}{4}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the factor $\frac{1}{2}$ → $\ln\left(\frac{13}{4}\right)$
• Used only the upper limit → $\dfrac{1}{2}\ln\left(13\right)$

Answer: $e^{2} + 1$

By parts with $u = x$ and $dv = e^{x}dx\mathclose{:}$ $\int xe^{x}dx = xe^{x} - \int e^{x}dx = xe^{x} - e^{x}\mathclose{.}$ From $0$ to $2$ that is $e^{2} + 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the second term of integration by parts → $2e^{2}$
• Got the sign of the second term wrong → $3e^{2} - 1$

Answer: $5e^{3} - 2$

By parts twice. With $u = x^2\mathclose{:}$ $\int x^2e^x dx = x^2e^x - \int 2xe^x dx\mathclose{;}$ with $u = 2x$ again: $\int 2xe^x dx = 2xe^x - 2e^x\mathclose{.}$ So the antiderivative is $e^x(x^2 - 2x + 2)\mathclose{,}$ and from $0$ to $3$ that is $5e^{3} - 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stopped after one integration by parts, dropping the integral left over → $9e^{3}$
• Added each integral instead of subtracting it → $17e^{3} - 2$

Answer: $6 + \ln(3)$

Divide first: $\dfrac{x^{2} + 3 x + 3}{x + 1} = x + 2 + \dfrac{1}{x + 1}\mathclose{.}$ So the integral is $\left[\dfrac{x^{2}}{2} + 2 x + \ln(x + 1)\right]_{0}^{2} = 6 + \ln(3)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Dropped the remainder after dividing → $6$
• Divided only the leading terms, keeping just $x$ → $2$

Answer: $124$

Put $u = x^{2} + 1\mathclose{,}$ so $du = 2x\,dx\mathclose{,}$ and the limits become $u = 1$ and $u = 5\mathclose{:}$ the integral is $\displaystyle\int_{1}^{5} 3u^{2}\,du = \left[\dfrac{3u^{3}}{3}\right]_{1}^{5} = 124\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to divide by the new power $3$ → $372$
• Kept the $x$-limits $0$ and $2$ for $u$ → $8$

Answer: $\frac{\pi}{2}$

Use the double-angle formula: $\cos^{2}(x) = \dfrac{1 + \cos(2x)}{2}\mathclose{.}$ Its antiderivative is $\dfrac{x}{2} + \dfrac{\sin(2x)}{4}\mathclose{,}$ and $\sin(2x)$ is $0$ at both limits, so the integral is $\dfrac{1}{2} \cdot \pi = \frac{\pi}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the $\frac{1}{2}$ of the double-angle formula → $\pi$
• Integrated $\cos^2(x)$ like a power, as $\frac{1}{3}\cos^3(x)$ → $- \frac{2}{3}$

Answer: $\dfrac{1}{4}\ln\left(\frac{5}{3}\right)$

Split into partial fractions: $\dfrac{1}{\left(x - 2\right)\left(x + 2\right)} = \dfrac{1}{4}\left(\dfrac{1}{x - 2} - \dfrac{1}{x + 2}\right)\mathclose{.}$ The antiderivative is $\dfrac{1}{4}\ln\left|\dfrac{x - 2}{x + 2}\right|\mathclose{;}$ from $3$ to $4$ that is $\dfrac{1}{4}\ln\left(\frac{5}{3}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot the factor $\dfrac{1}{4}$ from the partial fractions → $\ln\left(\frac{5}{3}\right)$
• Added the two logarithms instead of subtracting them → $\dfrac{1}{4}\ln\left(\frac{12}{5}\right)$

Answer: $\ln\left(\frac{5}{3}\right)$

$\int \dfrac{1}{x + 3}dx = \ln\left|x + 3\right|\mathclose{.}$ From $0$ to $2\mathclose{:}$ $\ln(5) - \ln(3) = \ln\left(\frac{5}{3}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used only the upper limit → $\ln\left(5\right)$
• Differentiated instead, using $-\dfrac{1}{(x + a)^2}$ → $\dfrac{16}{225}$

Answer: $-2$

By parts with $u = x$ and $dv = \cos(x)\,dx\mathclose{,}$ so $v = \sin(x)\mathclose{:}$ $\int x\cos(x)\,dx = x \cdot \sin(x) - \int \sin(x)\,dx = x\sin(x) + \cos(x)\mathclose{.}$ From $0$ to $\pi$ that is $-2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the second term of integration by parts → $0$
• Got the sign of the second term wrong → $2$

Answer: $\dfrac{e^{\frac{\pi}{2}} - 1}{2}$

By parts twice the integral comes back: call it $I\mathclose{,}$ and the two steps give $I = (\text{known terms}) - I\mathclose{,}$ so $2I$ is known. The antiderivative is $\dfrac{e^{x}\left(\sin(x) + \cos(x)\right)}{2}\mathclose{,}$ and from $0$ to $\frac{\pi}{2}$ that is $\dfrac{e^{\frac{\pi}{2}} - 1}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to divide by 2 once the integral came back → $e^{\frac{\pi}{2}} - 1$
• Used only the upper limit → $\dfrac{e^{\frac{\pi}{2}}}{2}$

Answer: $\dfrac{1}{2}\ln\left(\frac{6}{5}\right)$

The top is half the derivative of the bottom, so $\int \dfrac{x}{x^{2} + 5}dx = \dfrac{1}{2}\ln\left(x^{2} + 5\right)\mathclose{.}$ From $0$ to $1\mathclose{:}$ $\dfrac{1}{2}\ln(6) - \dfrac{1}{2}\ln(5) = \dfrac{1}{2}\ln\left(\frac{6}{5}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the factor $\frac{1}{2}$ → $\ln\left(\frac{6}{5}\right)$
• Used only the upper limit → $\dfrac{1}{2}\ln\left(6\right)$

Answer: $\dfrac{3e^{4} + 1}{4}$

By parts with $u = x$ and $dv = e^{2x}dx\mathclose{:}$ $\int xe^{2x}dx = \dfrac{xe^{2x}}{2} - \int \dfrac{e^{2x}}{2}dx = \dfrac{xe^{2x}}{2} - \dfrac{e^{2x}}{4}\mathclose{.}$ From $0$ to $2$ that is $\dfrac{3e^{4} + 1}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the second term of integration by parts → $e^{4}$
• Got the sign of the second term wrong → $\dfrac{5e^{4} - 1}{4}$

Answer: $2e^{2} - e$

By parts twice. With $u = x^2\mathclose{:}$ $\int x^2e^x dx = x^2e^x - \int 2xe^x dx\mathclose{;}$ with $u = 2x$ again: $\int 2xe^x dx = 2xe^x - 2e^x\mathclose{.}$ So the antiderivative is $e^x(x^2 - 2x + 2)\mathclose{,}$ and from $1$ to $2$ that is $2e^{2} - e\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stopped after one integration by parts, dropping the integral left over → $4e^{2} - e$
• Added each integral instead of subtracting it → $10e^{2} - 5e$

Answer: $- \frac{5}{2} + 9\ln(2)$

Divide first: $\dfrac{x^{2} - 2 x + 6}{x + 1} = x - 3 + \dfrac{9}{x + 1}\mathclose{.}$ So the integral is $\left[\dfrac{x^{2}}{2} - 3 x + 9\ln(x + 1)\right]_{0}^{1} = - \frac{5}{2} + 9\ln(2)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Dropped the remainder after dividing → $-\dfrac{5}{2}$
• Divided only the leading terms, keeping just $x$ → $\dfrac{1}{2}$

Answer: $\dfrac{248}{3}$

Put $u = x^{2} + 1\mathclose{,}$ so $du = 2x\,dx\mathclose{,}$ and the limits become $u = 1$ and $u = 5\mathclose{:}$ the integral is $\displaystyle\int_{1}^{5} 2u^{2}\,du = \left[\dfrac{2u^{3}}{3}\right]_{1}^{5} = \dfrac{248}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to divide by the new power $3$ → $248$
• Kept the $x$-limits $0$ and $2$ for $u$ → $\frac{16}{3}$

Answer: $\frac{3\pi}{4}$

Use the double-angle formula: $\cos^{2}(x) = \dfrac{1 + \cos(2x)}{2}\mathclose{.}$ Its antiderivative is $\dfrac{x}{2} + \dfrac{\sin(2x)}{4}\mathclose{,}$ and $\sin(2x)$ is $0$ at both limits, so the integral is $\dfrac{1}{2} \cdot \frac{3\pi}{2} = \frac{3\pi}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the $\frac{1}{2}$ of the double-angle formula → $\frac{3\pi}{2}$
• Integrated $\cos^2(x)$ like a power, as $\frac{1}{3}\cos^3(x)$ → $- \frac{1}{3}$

Answer: $\dfrac{1}{2}\ln\left(\frac{3}{2}\right)$

Split into partial fractions: $\dfrac{1}{\left(x - 1\right)\left(x + 1\right)} = \dfrac{1}{2}\left(\dfrac{1}{x - 1} - \dfrac{1}{x + 1}\right)\mathclose{.}$ The antiderivative is $\dfrac{1}{2}\ln\left|\dfrac{x - 1}{x + 1}\right|\mathclose{;}$ from $2$ to $3$ that is $\dfrac{1}{2}\ln\left(\frac{3}{2}\right)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot the factor $\dfrac{1}{2}$ from the partial fractions → $\ln\left(\frac{3}{2}\right)$
• Added the two logarithms instead of subtracting them → $\dfrac{1}{2}\ln\left(\frac{8}{3}\right)$