\(\displaystyle \int 2x\left(x^2 - 1\right)^4\,dx\)
$$\begin{align}
& u = x^2 - 1 \\
& du = 2x\,dx \\[1.75em]
& \int 2x\left(x^2 - 1\right)^4\,dx \\[0.4em]
=\ & \int u^4\,du \\[0.4em]
=\ & \dfrac{u^5}{5} + C \\[0.4em]
=\ & \dfrac{\left(x^2 - 1\right)^5}{5} + C
\end{align}$$
\(\displaystyle \int 3x^2\left(x^3 + 5\right)^7\,dx\)
$$\begin{align}
& u = x^3 + 5 \\
& du = 3x\,dx \\[1.75em]
& \int 3x^2\left(x^3 + 5\right)^7\,dx \\[0.4em]
=\ & \int u^7\,du \\[0.4em]
=\ & \dfrac{u^8}{8} + C \\[0.4em]
=\ & \dfrac{\left(x^3 + 5\right)^8}{8} + C
\end{align}$$
\(\displaystyle \int \dfrac{x^3}{\sqrt{1 - x^4}}\,dx\)
\(\dfrac{-1}{2}\left(1 - x^4\right)^{1 / 2} + C\)
\(\displaystyle \int x\sqrt{x + 2}\,dx\)
\(\dfrac{2}{5}(x + 2)^{5 / 2} - \dfrac{4}{3}(x + 2)^{3 / 2} + C\)
\(\displaystyle \int \dfrac{\sin x}{\cos^3 x}\,dx\)
\(\dfrac{1}{2\cos^2 x} + C\)
\(\displaystyle \int \left(x^2 + 1\right)^2(2x)\,dx\)
\(\dfrac{\left(x^2 + 1\right)^3}{3} + C\)
\(\displaystyle \int \left(x^3 + 1\right)^4\left(3x^2\right)\,dx\)
\(\dfrac{\left(x^3 + 1\right)^5}{5} + C\)
\(\displaystyle \int 3^{\sin \theta}\cos \theta\,d\theta\)
\(\dfrac{3^{\sin \theta}}{\ln 3} + C\)
\(\displaystyle \int \left(x^2 - 5\right)^3x\,dx\)
\(\dfrac{1}{8}\left(x^2 - 5\right)^4 + C\)
\(\displaystyle \int \left(x^3 - 3\right)^2\left(4x^2\right)\,dx\)
\(\dfrac{4}{9}\left(x^3 - 3\right)^3 + C\)
\(\displaystyle \int \sqrt{5x + 3}\,dx\)
\(\dfrac{2}{15}(5x + 3)^{3 / 2} + C\)
\(\displaystyle \int x\sqrt{5x + 3}\,dx\)
\(\dfrac{1}{25}\left(\dfrac{2(5x + 3)^{5 / 2}}{5} - 2(5x + 3)^{3 / 2}\right) + C\)
\(\displaystyle \int \sin^2(4x)\cos(4x)\,dx\)
\(\dfrac{1}{12}\sin^3 4x + C\)
\(\displaystyle \int_0^2 \dfrac{-4x}{\sqrt{9 - 2x^2}}\,dx\)
\(-4\)
\(\displaystyle \int e^{-5x}\,dx\)
\(\dfrac{-1}{5}e^{-5x} + C\)
Integrating Exponential Functions By Substitution - Antiderivatives - Calculus
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\(\displaystyle \int x^3e^{x^4}\,dx\)
\(\dfrac{1}{4}e^{x^4} + C\)
Integrating Exponential Functions By Substitution - Antiderivatives - Calculus
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\(\displaystyle \int e^x\sqrt{1 - e^x}\,dx\)
\(\dfrac{-2}{3}\left(1 - e^x\right)^{3 / 2} + C\)
Integrating Exponential Functions By Substitution - Antiderivatives - Calculus
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\(\displaystyle \int \dfrac{e^x + e^{-x}}{e^x - e^{-x}}\,dx\)
\(\ln \left\lvert e^x - e^{-x} \right\rvert + C\)
Integrating Exponential Functions By Substitution - Antiderivatives - Calculus
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\(\displaystyle \int \dfrac{e^{1 / x^2}}{x^3}\,dx\)
\(\dfrac{-1}{2}e^{1 / x^2} + C\)
Integrating Exponential Functions By Substitution - Antiderivatives - Calculus
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\(\displaystyle \int \dfrac{e^{3x} + 4e^x + 5}{e^x}\,dx\)
\(\dfrac{e^{2x}}{2} + 4x + \dfrac{5e^{-x}}{-1} + C\)
Integrating Exponential Functions By Substitution - Antiderivatives - Calculus
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\(\displaystyle \int 6e^{6x}\,dx\)
\(e^{6x} + C\)
\(\displaystyle \int 2xe^{x^2}\,dx\)
\(e^{x^2} + C\)
\(\displaystyle \int \dfrac{e^x + 1}{e^x}\,dx\)
\(x - e^{-x} + C\)