How do you evaluate $\int \frac{x}{\sqrt{1 + x^{2}}}$ $int x/sqrt(1+x^2)$ from $[- \sqrt{2}, 0]$ $[-sqrt2,0]$ ?

Question

How do you find the integral $\int \frac{1}{x^{2} \cdot \sqrt{x^{2} - 9}} d x$ $int1/(x^2*sqrt(x^2-9))dx$ ?
How do you find the integral $\int \frac{x^{3}}{\sqrt{x^{2} + 9}} d x$ $intx^3/(sqrt(x^2+9))dx$ ?
How do you find the integral $\int x^{3} \cdot \sqrt{9 - x^{2}} d x$ $intx^3*sqrt(9-x^2)dx$ ?
How do you find the integral $\int \frac{x^{3}}{\sqrt{16 - x^{2}}} d x$ $intx^3/(sqrt(16-x^2))dx$ ?
How do you find the integral $\int \frac{\sqrt{x^{2} - 1}}{x} d x$ $intsqrt(x^2-1)/xdx$ ?
How do you find the integral $\int \frac{\sqrt{x^{2} - 9}}{x^{3}} d x$ $intsqrt(x^2-9)/x^3dx$ ?
How do you find the integral $\int \frac{x}{\sqrt{x^{2} + x + 1}} d x$ $intx/(sqrt(x^2+x+1))dx$ ?
How do you find the integral $\int \frac{d t}{\sqrt{t^{2} - 6 t + 13}}$ $intdt/(sqrt(t^2-6t+13))$ ?
How do you find the integral $\int x \cdot \sqrt{1 - x^{4}} d x$ $intx*sqrt(1-x^4)dx$ ?
How do you prove the integral formula $\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = ln (x + \sqrt{x^{2} + a^{2}}) + C$ $intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C$ ?

1583 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-26T16:34:57

$1 - \sqrt{3}$ $1-sqrt3$

$\int_{- \sqrt{2}}^{0} \frac{x}{\sqrt{1 + x^{2}}} d x$ $int_-sqrt2^0x/sqrt(1+x^2)dx$

$\int_{- \sqrt{2}}^{0} x {(1 + x^{2})}^{- \frac{1}{2}} d x$ $int_-sqrt2^0x(1+x^2)^(-1/2)dx$

by inspection

$= {[{(1 + x^{2})}^{\frac{1}{2}}]}_{- \sqrt{2}}^{\frac{1}{2}}$ $=[(1+x^2)^(1/2)]_-sqrt2^0$

$= {[{(1 + x^{2})}^{\frac{1}{2}}]}^{0} - {[{(1 + x^{2})}^{\frac{1}{2}}]}_{- \sqrt{2}}^{\frac{1}{2}}$ $=[(1+x^2)^(1/2)]^0-[(1+x^2)^(1/2)]_-sqrt2$

$= 1 - {(1 + 2)}^{\frac{1}{2}}$ $=1-(1+2)^(1/2)$

$= 1 - \sqrt{3}$ $=1-sqrt3$

How do you evaluate ∫x√1+x2int x/sqrt(1+x^2) from [−√2,0][-sqrt2,0]?