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It's xn+1n+1+C\frac{x^{n+1}}{n+1} + Cn+1xn+1+C, for n≠−1n \neq -1n=−1.
The derivative is du/dx=2xdu/dx = 2xdu/dx=2x.
dx=du2xdx = \frac{du}{2x}dx=2xdu after differentiating.
It's 13(1+x2)3/2+C\frac{1}{3}(1 + x^2)^{3/2} + C31(1+x2)3/2+C, via substitution.
∫u1/2du=23u3/2+C\int u^{1/2} du = \frac{2}{3} u^{3/2} + C∫u1/2du=32u3/2+C.
13(1+x2)3/2+C\frac{1}{3} (1 + x^2)^{3/2} + C31(1+x2)3/2+C
Let u=1+cos(2x)u = 1 + \cos(2x)u=1+cos(2x) for substitution.
Use x=sec(θ)x = \sec(\theta)x=sec(θ) for substitution.
dx=sec(θ)tan(θ)dθdx = \sec(\theta)\tan(\theta) d\thetadx=sec(θ)tan(θ)dθ.
It simplifies to (ln(x))22+C\frac{(\ln(x))^2}{2} + C2(ln(x))2+C.
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