Indefinite integration practice · Mixed difficulty

Option A ($-2 e^{\tan ^2 x} \cos ^4 x+c$) is correct because after substitution and integration, the result matches this form.

Option A is correct because the integral of (1-x²)log(x) dx is found using integration by parts, resulting in the given expression.

Option B ($-\cos x-x \sin x+C$) is correct because it represents the antiderivative of the function $f(x)=x|\cos x|$ in the given interval.

Option C ($a=\pi / 4, c$ is arbitrary constant) is correct because the integral of the given expression results in a sine function with a phase shift of $\pi/4$ and an arbitrary constant of integration.

Option B is correct. The integral is solved using substitution, where $t = 3 + 4\cos x$, simplifying the expression and allowing for direct integration.

Option B $\left( {\frac{1}{5},0} \right)$ is correct because by using the reduction formula for the integral of tan^n(x), we can find the values of a and b.

Option C ($f(x) g^{\prime}(x)-f^{\prime}(x) g(x)$) is correct because it results from applying integration by parts twice to the given integral.

Option B (\$1/2)(\log(x + \sqrt{1+x^2}))^2 + C$ is correct. The integral is solved by using the substitution \$t = \log(x + \sqrt{1+x^2}))$.

Option C (7) is correct because after solving the integral and comparing coefficients, we find m=2 and n=3, thus 2m+n=7.

Option C is correct because the integral is solved using a substitution method, resulting in the given expression with a fractional exponent.

Option B ($2\,\,{\tan ^{ - 1}}\,\frac{{\sqrt {{{{x^2} + \,\,x\,\, + 1}}} }}{x}$) is correct because the integral simplifies to this expression after substitution and integration.

Option C ($\frac{1}{3} x^3 \phi(x^3)-\int x^2 \phi(x^3) d x+C$) is correct because using substitution and integration by parts leads to the correct expression.

Option D (None of these) is correct because the provided solution appears to have an error in the initial problem statement and substitution, leading to an incorrect result. The question asks for a definite integral, but the provided solution calculates an indefinite integral.

Option B ($\frac{1}{2}\left[\tan ^{-1} x^2\right]^2+c$) is correct because it is the result of a u-substitution and applying the power rule for integration.

Option B ($\log _e(1+\sqrt{2})-\frac{\pi}{4}$) is correct because after simplifying the integral and applying the initial condition $f(0)=0$, we find $f(1) = \log_e(1+\sqrt{2}) - \frac{\pi}{4}$.

Option D is correct because the integral of x sin(kx) dx is found using integration by parts, resulting in -x cos(kx)/k + sin(kx)/k² + c.

Option B ($\frac{{1}}{{3}} \log |\frac{{\sqrt{{1-x^3}}-1}}{{\sqrt{{1-x^2}}+1}}| + C$) is correct because it is the result of using substitution and partial fractions to evaluate the integral.

Option D ($\tan \,\left( {x\\;{e^x}} \right)$) is correct because by substituting $t = xe^x$, the integral simplifies and can be easily solved.

Option B ($\frac{{16}}{5}$) is correct because after applying integration by parts and evaluating the definite integral, the result is 16/5.

Option D ($x e^{x+x^{-1}}+C$) is correct because it results from integrating by parts and simplifying the expression.

Option C ($f(x)=-(1 / 3)\left[\left(1-x^2\right)^{3 / 2}-8\right]$) is correct because integrating $x \sqrt{1-x^2}$ and applying the initial condition $f(0) = 7/3$ yields the correct function.

Option C (- log ,| e ^{- x} + ,√e ^{- 2x} - , 1 | , + ,\c$- \log \\,\left| {{{{e^{ - x}}}\\ + \\,\sqrt {{{{e^{ - 2x}}} - \\,\\1} } } \right|\\,\\ + \\,\c$) is correct because it is the result of integrating the given expression using substitution.

Option C ($x + \log \,\left( {\frac{{x - 1}}{{x + 1}}} \right) + c$) is correct because after performing the integration, the result matches this expression.

The correct answer is $e^x(\tan ^{-1} x-\frac{1}{1+x^2})+c$ because it fits the form of the integral $\int e^x(f(x)+f'(x))dx = e^x f(x) + c$.

Option A (${\sin ^{ - 1}}\,\,\left( {2x\, - \,3} \right)\,\,+ \,c$) is correct because it is the result of completing the square and applying the standard integral formula for arcsine.

Option B ($\frac{1}{4}\log \left( {{x^4} + 1} \right) + C$) is correct because it is the result of evaluating the indefinite integral by simplifying the expression and using a simple substitution.

Option B ($\frac{1}{3}\\ {\tan ^{ - 1}}\left( {{x^3}} \right) + c$) is correct because by substituting $t = x^3$, the integral simplifies to a standard arctangent integral.

Option A ($\tan \frac{x}{2}\\ + \\ c$) is correct because the integral simplifies to $\int \frac{1}{{1 + \cos x}} dx$, which evaluates to $\tan \frac{x}{2}\\ + \\ c$.

Option D (1) is correct because, after solving the integral and comparing coefficients, the absolute difference between m and n is 1.

Option B ($\mathrm{x}^4\{8(\log x)^2-4(\log \mathrm{x})+1\}+\mathrm{C}$) is correct because it is the result of integrating the given expression using integration by parts twice.